# NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1

Mathematics is an important subject in the CBSE curriculum because every student must have a good understanding of Mathematics. The NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1  cover an Introduction to Class 11 Maths Chapter 4 Exercise 4.1. An introduction must be included in the solutions as it is based on concepts. Numerical values are of no consequence if the concept behind a particular topic in Mathematics is not understood correctly. The introduction has been prepared by experts on the subject and years of experience have been put into it. The introduction to the subject, might be mistaken for a synopsis of the subject matter, but it is not so. Especially in Mathematics, people spend a lot of time meticulously going through the subject to gather and understand the whole topic before using the information in a very precise manner to cover the whole topic in a very concise form.

The introduction gives an insight into the details which are discussed in the topic along with the concept. The concepts are very important for students to understand because the whole understanding of the chapter depends on them. An elaborative understanding of Chapter 4 concepts is covered in the NCERT Solutions Class 11 Maths Chapter 4 Exercise 4.1.  Students might need help understandingthe concepts of Mathematics. There are many concepts present in Mathematics. The NCERT Solutions provide explanations and cover all of the topics. The range of solutions starts from NCERT Solutions Class 1 to NCERT Solutions Class 12. Students can access and download all of these topics from the Extramarks website.This enables them to access them offline whenever required. These solutions are a tool that is always available for the students to access to clarify their doubts. The NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1 is also available on the Extramarks’ website in PDF format.

The concepts for the Principle of Mathematical Induction have been discussed in the NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1  which is an integral part of Mathematics. Examples of the concepts are also present in the NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1. These examples are helpful for the students to understand the concepts and their applications. In total, there are six samples, which have been discussed in the Chapter. The introduction part has been designed to clarify the concept of Mathematics which has been discussed in the Chapter. Once the students have understood the concepts discussed in the Chapter, they will be able to solve all  twenty-four questions that have been set for this Chapter. These NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1 can be easily downloaded from the Extramarks’ website. The Class 11 Maths Chapter 4 Exercise 4.1 Solution has been organised in a logical order for easy comprehension.It is important to note that the question paper might include at least one of the questions discussed in the Chapter. The NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1  has all of them covered. The NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1 have been designed in such a way that they not only explain the concepts and the calculations of the topics covered in them but also provide an explanation of the reasons the concepts and calculations are being used. If the subject is not explained properly, the students will have a huge difficulty  understanding the topic. The teachers’ explanation in the NCERT Solutions for Class 11 Maths, Chapter 4, Exercise 4.1, is detailed and simple to understand.The method in which the chapter has been explained in the NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1 has been formulated after a lot of research, and the explanations have been tailored to the capacity of the students to understand. The NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1 can be accessed via the Extramarks’ website.

Since Mathematics is the study of numerical concepts, the students must try to solve as many problems as possible to understand whether they are successful in understanding the concept. This, in turn, develops an interest in the subject. Mastering a subject requires a strong interest in it.Human psychology is such that if a student is successful in understanding a topic, the student will make efforts to conquer further problems and will be able to master the subject further. However, if the student is not able to understand the topic or is not able to clarify doubts which lead to further misunderstandings about the subject, the student loses interest in the subject as a whole.  The NCERT Solutions for Class 11 Maths, Chapter 4, Exercise 4.1, teaches students how to avoid such situations.

Step-by-step explanations have been provided in the NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1 for the students. The more a student practises the solutions provided in the NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1  increases the chances of scoring higher marks are obtained. These solutions not only help students to secure higher marks but also enhances their knowledge base. Some students achieve a higher goal in life and are successful in becoming engineers and scientists. Aspiration for higher study is very important in a student. The NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1 have been designed to encourage students to proceed a step further with their exam preparation. The NCERT Solutions for Class 11 Maths Chapter 4 Exercise 4.1 is a learning tool designed to boost the exam preparation of the students.

## NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction (Ex 4.1) Exercise 4.1

A student should practise the exercises with the help of NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1 to understand the topic. The solutions for this chapter have been provided as a PDF file so that students can download and use them offline as needed.What will You Learn in Exercise 4.1 of NCERT Class 11 Maths Chapter 4?

There is a list of concepts discussed in the chapter provided in NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1  to enable the students to learn and understand concepts and to clarify their doubts.

Students have to solve problems like proving the following to be true by using Mathematical Induction for all n where n stands for Natural numbers. Since n is not a constant in the problem, it can be any number, it has to be proven by using the concept of Mathematical Induction that the variable ‘n’ is a Natural Number.

Natural numbers have been discussed in previous classes. A student can find the mention of Natural numbers in NCERT Solutions Class 8,  NCERT Solutions Class 9, and NCERT Solutions Class 10.  The topic of Natural Numbers has been discussed in detail in these classes. Students can find the explanations in the NCERT Solutions. Students can always access Extramarks’ website to obtain Solutions. Students are made to practise the basics until they become strong at it. Doubts can come up while understanding a particular concept. The NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1  are designed in such a manner, that a student can refer to the basics and rectify any doubts that are persistent. NCERT always tries to give a standard level of understanding to the students of a certain class after a lot of research. The NCERT Solutions for Class 11 Maths, Chapter 4, Exercise 4.1, explains the level of preparation needed.The time that is taken to understand the basics of a particular topic is taken into consideration. The method of introduction is also decided after extensive research and consideration of the student’s age and ability to understand.he NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1  which is accessible through the Extramarks’ website, has been built under the strict guidelines of NCERT, so it may be beneficial to the students who access it and use it.

Teachers suggest that students should try to solve the problems on their own before they try to find solutions in the NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1. Since all the questions in the chapter belong to the same topic, most of the twenty-four questions will be the same as one another.

The questions are intended to allow students to repeatedly practise the concepts and calculationsThis not only helps the students to understand but also enables them to become familiar with the calculations so that they do not have to spend a lot of time calculating and solving such problems. Repeated practise enables the students to perform the calculations almost as if by habit. This ensures that the students gain the ability to solve higher-level problems with easeSince the topic discussed in the NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1  is only a part of the subject, a student is expected not to spend a lot of time with calculations. This makes the students fast and accurate when the Principle of Mathematical Induction is used as a part of a bigger problem to solve or  is integrated into a bigger problem presented to solve in a higher Class. Mathematics is a subject which has a huge number of principles, concepts, rules, formulas,  riders, etc. It might sometimes be confusing, and students might tend to avoid the subject as a whole. But this subject is fun for the students who give it time and perseverance. Once a student starts to understand the concepts and rules, it becomes a habit for the student to learn, understand, and solve problems mathematically. This is a habit that pushes the students to take on further challenges to solve more and more problems and get the concepts clear in their minds. Sometimes students become so engrossed with Mathematics that they start deriving fun from calculating and solving new problems. Students begin to take on new challenges and do not want to repeat the same problems over and over.The NCERT Solutions for Class 11 Maths Chapter 4 Exercise 4.1 assist students with a part of the topic.

Since students deal with Natural Numbers in the chapter, the first Natural Number will automatically be the choice to take into consideration, that is 1. Using the Principle of Induction, it has to be proven that the product of each calculation will be the next Natural Number. A chain of Natural Numbers develops as a result. The equation that is formed by using the Principle of Induction has two sides, the left-hand side and the right-hand side. The student has to prove the Right-hand side to be true by solving the Left-hand side.

The problems mentioned in the NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1  are fun to solve if the student understands the concepts. It is advisable to avoid short form and make it a habit, as this helps students  fetch better marks even in higher studies, like at the graduation level or engineering studies. The sum of all Natural Numbers also comes into action, which has been discussed in previous classes to solve equations. The NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1 provides details of this aspect of the topic. The Chapter discusses the concept to derive the sum of all Natural Numbers. As the student continues to practice, the topic becomes simple. Many important questions have been included in the NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1  to enable the students to prepare for examinations.

The application of a few riders has been included in the chapter which the students have to use to solve the problems. The student is supposed to understand and learn the required riders in the previous classes before reaching this level of the subject. Explanations have been provided in the NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1. Students can access the solutions and practise along with the explanations to master the topic.

The NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1 have been designed for a particular Chapter. The NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1 is an example of the solutions provided by the NCERT. These NCERT Solutions include the solutions to Chapter 4 that are provided on the Extramarks’ website. The NCERT Solutions for Class 11 Maths, Chapter 4, Exercise 4.1 is one of many helpful resources for students.Access NCERT Solutions for Class 11 Mathematics Chapter 4 – Principle of Mathematics Induction

The Principle of Mathematics Induction has been discussed in the NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1 in detail and these solutions are available at the Extramarks’ website where the students can access the solutions, study, and practise them to clarify doubts and understand the concept of the Principle of Mathematical Induction. The NCERT Solutions are compiled in an easy to understand language and can help students with their overall preparation for the board exams.

Mathematics begins at the very first level, when a student begins to attend school.The concept of Mathematics is revealed step by step to the students, so they can understand them with time. If students have questions about previous classes, they can always review the study material from NCERT Solutions Class 1 and Class 2, or NCERT Solutions Class 3 or Class 4, and so on.Solutions for NCERT Solutions Class 5, NCERT Solutions Class 6, and NCERT Solutions Class 7 are also available on Extramarks’ website for the students to access and clarify doubts. This website is extremely beneficial to students because textbooks from lower classes are frequently discarded as students advance to higher classes.Finding textbooks in lower classes can be difficult for students at times.Complicated concepts are revealed to students as they complete higher standards.Sometimes a few concepts may take more time to understand and detailed explanations are required. NCERT Solutions Class 8, NCERT Solutions Class 9, and NCERT Solutions Class 10 are available on the Extramarks’ website to be accessed. NCERT Solutions Class 11 and NCERT Solutions Class 12, are available to assist students with the concepts of Mathematics if they require some extra time to understand them.

### NCERT Solutions for Class 11 Maths Chapters

There are several topics included in the Chapters of Mathematics for Class 11. Sets, Relations and Functions, Trigonometric Functions, Principle of Mathematical Induction, Complex numbers and Quadratic Equations are examples of the topic covered in Class 11 Maths. All these topics have been incorporated in the NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1.

The NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1 has all the topics of Chapter 4 covered. The NCERT Solutions for Class 11 Maths Chapter 4 Exercise 4.1 are the complete solutions to boost the students’ preparation.

As we know Trigonometry deals with the concept of angles. Angles are formed in between the Initial side and the Terminal side, the angles can be measured by degrees, minutes and seconds. Radian Measure Concept has also been included. Sin and Cos are the calculations which are done on the Base, Hypotenuse and Tangent of a Triangle.  The Principle of Mathematical Induction has been discussed in detail in the NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1. Along with the solutions for Class 11, there are many other topics  covered in the NCERT Solutions which help the students further. The NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1 have been provided only for the specific Chapter and Exercise. The NCERT Solutions for Class 11 Maths Chapter 4 Exercise 4.1 are compiled following the guidelines issued by the CBSE.

8

The Quadratic formula has also been discussed in detail. The solutions for Exercise 4.1 have all these topics covered. Calculation of discriminants, Real route or route or iota. Closure law, Associative Law and Commutative Law, Multiplicative identity, Multiplicative Inverse. Distributive Law, Modules of Complex numbers, Conjugate Complex numbers, and Polar Representation of Complex Numbers etc are the topics that have been discussed in the Chapters. The NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1 provides the solutions for the students.

NCERT Solutions for Class 11 Maths ChapteCr 4 Principle of Mathematical Induction Exercise 4.1

Under the Principle of Mathematical Induction Exercise 4.1. The Principle of Mathematical Induction and its utility have been discussed and Exercise 4.1 provides problems relating to the utility of the Principle of Mathematical Induction. Students can practise solving the problems in order to get used to the concept and the formulas used to solve them. The NCERT Solutions for Class 11 Maths Chapter 4 Exercise 4.1 havean abundance of solutions which can be used by the students for practice. Keeping in mind that the topic has also been discussed in Class 10, it is easier for students to understand the Principle of Mathematical Induction in Class 11. Students who are eager to pursue Engineering to Graduate have to be well versed in the topic. The NCERT Solutions for Class 11 Maths Chapter 4 Exercise 4.1 is available on the Extramarks’ website and is available to students whenever required. The NCERT Solutions for Class 11 Maths Chapter 4 Exercise 4.1 can be relied upon during  final exam preparation.

Q.1 Prove by using principal of mathematical induction for all n ∈ N

$\mathbf{1}\mathbf{+}\mathbf{3}\mathbf{+}{\mathbf{3}}^{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}{{\mathbf{3}}^{\mathbf{n}}}^{\mathbf{–}\mathbf{1}}\mathbf{=}\frac{\mathbf{\left(}{\mathbf{3}}^{\mathbf{n}}\mathbf{-}\mathbf{1}\mathbf{\right)}}{\mathbf{2}}\mathbf{.}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{​ be the given statement,}\\ \mathrm{i}.\mathrm{e}.,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{n}\right):1+3+{3}^{2}+...+{{3}^{\mathrm{n}}}^{–1}=\frac{\left({3}^{\mathrm{n}}-1\right)}{2}\\ \mathrm{For}\text{n}=1,\\ \mathrm{P}\left(1\right)\text{\hspace{0.17em}\hspace{0.17em}}=1\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{{3}^{1}-1}{2}=1=\mathrm{L}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k,}\\ \text{i.e., P}\left(\mathrm{k}\right):1+3+{3}^{2}+...+{{3}^{\mathrm{k}}}^{–1}=\frac{\left({3}^{\mathrm{k}}-1\right)}{2}...\left(1\right)\\ \text{We need to prove that P(k + 1) is true whenever P(k) is true.}\\ \text{We have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1+3+{3}^{2}+...+{{3}^{\mathrm{k}}}^{–1}+{3}^{\mathrm{k}}=\frac{\left({3}^{\mathrm{k}+1}-1\right)}{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\left({3}^{\mathrm{k}}-1\right)}{2}+{3}^{\mathrm{k}}=\frac{\left({3}^{\mathrm{k}+1}-1\right)}{2}...\left(2\right)\left[\mathrm{From}\text{equation}\left(1\right)\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\left({3}^{\mathrm{k}}-1\right)}{2}+{3}^{\mathrm{k}}=\frac{\left({3}^{\mathrm{k}}-1+2{.3}^{\mathrm{k}}\right)}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3{.3}^{\mathrm{k}}-1}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{3}^{\mathrm{k}+1}-1}{2}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

Q.2 Prove by using principal of mathematical induction for all n ∈ N

${\mathbf{1}}^{\mathbf{3}}\mathbf{+}{\mathbf{2}}^{\mathbf{3}}\mathbf{+}{\mathbf{3}}^{\mathbf{3}}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}{\mathbf{n}}^{\mathbf{3}}\mathbf{=}{\mathbf{\left\{}\frac{\mathbf{n}\mathbf{\left(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{\right)}}{\mathbf{2}}\mathbf{\right\}}}^{\mathbf{2}}\mathbf{.}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{​ be the given statement,}\\ \mathrm{i}.\mathrm{e}.,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{n}\right):{1}^{3}+{2}^{3}+{3}^{3}+...+{\mathrm{n}}^{3}={\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right\}}^{2}\\ \mathrm{For}\text{n}=1,\\ \mathrm{P}\left(1\right)={1}^{3}=1\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:{\left\{\frac{1.\left(1+1\right)}{2}\right\}}^{2}=1.\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k,}\\ \text{i.e., P}\left(\mathrm{k}\right):{1}^{3}+{2}^{3}+{3}^{3}+...++{\mathrm{n}}^{\mathrm{k}}={\left\{\frac{\mathrm{k}\left(\mathrm{k}+1\right)}{2}\right\}}^{2}...\left(\mathrm{i}\right)\\ \text{We need to prove that}\mathrm{P}\left(\mathrm{k}+1\right)\text{is true whenever P(k) is true.}\\ \text{We have}\\ \text{\hspace{0.17em}\hspace{0.17em}}{1}^{3}+{2}^{3}+{3}^{3}+...++{\mathrm{k}}^{3}+{\left(\mathrm{k}+1\right)}^{3}={\left\{\frac{\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)}{2}\right\}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left\{\frac{\mathrm{k}\left(\mathrm{k}+1\right)}{2}\right\}}^{2}+{\left(\mathrm{k}+1\right)}^{3}={\left\{\frac{\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)}{2}\right\}}^{2}\text{\hspace{0.17em}}...\left(2\right)\\ \left[\mathrm{From}\text{equation}\left(\mathrm{i}\right)\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left\{\frac{\mathrm{k}\left(\mathrm{k}+1\right)}{2}\right\}}^{2}+{\left(\mathrm{k}+1\right)}^{3}={\left(\mathrm{k}+1\right)}^{2}\left\{\frac{\mathrm{k}}{4}+\mathrm{k}+1\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left(\mathrm{k}+1\right)}^{2}\left\{\frac{{\mathrm{k}}^{2}+4\mathrm{k}+4}{4}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left(\mathrm{k}+1\right)}^{2}\left\{\frac{{\left(\mathrm{k}+2\right)}^{2}}{4}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left\{\frac{\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)}{2}\right\}}^{2}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by principle}\\ \text{of mathematical induction, P(n) is true for every positive integer n.}\end{array}$

Q.3 Prove by using principal of mathematical induction for all n ∈ N

$\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{\left(}\mathbf{1}\mathbf{+}\mathbf{2}\mathbf{\right)}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{\left(}\mathbf{1}\mathbf{+}\mathbf{2}\mathbf{+}\mathbf{3}\mathbf{\right)}}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}\frac{\mathbf{1}}{\mathbf{\left(}\mathbf{1}\mathbf{+}\mathbf{2}\mathbf{+}\mathbf{3}\mathbf{+}\mathbf{4}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}\mathbf{n}\mathbf{\right)}}\mathbf{=}\frac{\mathbf{2}\mathbf{n}}{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{.}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{​ be the given statement,}\\ \mathrm{i}.\mathrm{e}.,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{n}\right):1+\frac{1}{\left(1+2\right)}+\frac{1}{\left(1+2+3\right)}+...+\frac{1}{\left(1+2+3+4+...+\mathrm{n}\right)}=\frac{2\mathrm{n}}{\mathrm{n}+1}\\ \mathrm{For}\text{n}=1,\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}}⇒\mathrm{P}\left(1\right)=1\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:\frac{2\left(1\right)}{1+1}=1.\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k,}\\ \text{i.e., P}\left(\mathrm{k}\right):1+\frac{1}{\left(1+2\right)}+\frac{1}{\left(1+2+3\right)}+...+\frac{1}{\left(1+2+3+4+...+\mathrm{k}\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2\mathrm{k}}{\mathrm{k}+1}...\left(1\right)\\ \text{We need to prove that}\mathrm{P}\left(\mathrm{k}+1\right)\text{is true whenever P(k) is true.}\\ \text{We have}\\ 1+\frac{1}{\left(1+2\right)}+\frac{1}{\left(1+2+3\right)}+...+\frac{1}{\left(1+2+3+4+...+\mathrm{k}\right)}+\\ \frac{1}{\left\{1+2+3+4+...+\left(\mathrm{k}+1\right)\right\}}=\frac{2\left(\mathrm{k}+1\right)}{\mathrm{k}+2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{2\mathrm{k}}{\mathrm{k}+1}+\frac{1}{\left\{1+2+3+4+...+\left(\mathrm{k}+1\right)\right\}}=\frac{2\left(\mathrm{k}+1\right)}{\mathrm{k}+2}...\left(2\right)\\ \left[\mathrm{From}\text{equation}\left(1\right)\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:\text{\hspace{0.17em}}\frac{2\mathrm{k}}{\mathrm{k}+1}+\frac{1}{\left\{1+2+3+4+...+\left(\mathrm{k}+1\right)\right\}}\text{\hspace{0.17em}}=\frac{2\mathrm{k}}{\mathrm{k}+1}+\frac{1}{\left\{\frac{\left(\mathrm{k}+1\right)\left(\mathrm{k}+1+1\right)}{2}\right\}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2\mathrm{k}}{\mathrm{k}+1}+\frac{2}{\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2}{\mathrm{k}+1}\left(\mathrm{k}+\frac{2}{\mathrm{k}+2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2}{\mathrm{k}+1}\left(\frac{{\mathrm{k}}^{2}+2\mathrm{k}+1}{\mathrm{k}+2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2}{\mathrm{k}+1}×\frac{{\left(\mathrm{k}+1\right)}^{2}}{\mathrm{k}+2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2\left(\mathrm{k}+1\right)}{\left(\mathrm{k}+1\right)+1}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

Q.4 Prove by using principal of mathematical induction for all n ∈ N

$\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{.}\mathbf{4}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}\mathbf{n}\mathbf{\left(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{\right)}\mathbf{\left(}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{\right)}\mathbf{=}\frac{\mathbf{n}\mathbf{\left(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{\right)}\mathbf{\left(}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{\right)}\mathbf{\left(}\mathbf{n}\mathbf{+}\mathbf{3}\mathbf{\right)}}{\mathbf{4}}\mathbf{.}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{​ be the given statement,}\\ \mathrm{i}.\mathrm{e}.,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{n}\right):1.2.3+2.3.4+...+\mathrm{n}\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)=\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)\left(\mathrm{n}+3\right)}{4}\\ \mathrm{For}\text{n}=1,\\ \mathrm{P}\left(1\right)=1.2.3=6\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1\left(1+1\right)\left(1+2\right)\left(1+3\right)}{4}=\frac{1\left(2\right)\left(3\right)\left(4\right)}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=6\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k,}\\ \text{i.e., P}\left(\mathrm{k}\right):1.2.3+2.3.4+...+\mathrm{k}\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\\ =\frac{\mathrm{k}\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)}{4}...\left(1\right)\\ \text{We need to prove that}\mathrm{P}\left(\mathrm{k}+1\right)\text{is true whenever P(k) is true.}\\ \text{We have}\\ 1.2.3+2.3.4+...+\mathrm{k}\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\text{\hspace{0.17em}}+\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)\\ =\frac{\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)\left(\mathrm{k}+4\right)}{4}\\ ⇒\frac{\mathrm{k}\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)}{4}+\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)\\ =\frac{\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)\left(\mathrm{k}+4\right)}{4}...\left(2\right)\\ \left[\mathrm{From}\text{equation}\left(1\right)\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{k}\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)}{4}+\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)\\ =\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)\left(\frac{\mathrm{k}}{4}+1\right)\text{\hspace{0.17em}}\\ =\frac{\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)\left(\mathrm{k}+4\right)}{4}\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

Q.5 Prove by using principal of mathematical induction for all n ∈ N

$\mathbf{1}\mathbf{.3}\mathbf{+}\mathbf{2}{\mathbf{.3}}^{\mathbf{2}}\mathbf{+}\mathbf{3}{\mathbf{.3}}^{\mathbf{3}}\mathbf{+}\mathbf{.}\mathbf{..}{\mathbf{\text{n.3}}}^{\mathbf{\text{n}}}\mathbf{=}\frac{\left(2\text{n}–1\right){\mathbf{3}}^{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{+}\mathbf{3}}{\mathbf{4}}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{​ be the given statement,}\\ \mathrm{i}.\mathrm{e}.,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{n}\right):1.3+2{.3}^{2}+3{.3}^{3}+...+\mathrm{n}.{3}^{\mathrm{n}}=\frac{\left(2\mathrm{n}-1\right){3}^{\mathrm{n}+1}+3}{4}\\ \mathrm{For}\text{n}=1,\\ \mathrm{P}\left(1\right)=1.3=3\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:\text{\hspace{0.17em}\hspace{0.17em}}\frac{\left(2×1-1\right){3}^{1+1}+3}{4}=\frac{12}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k,}\\ \text{i.e., P}\left(\mathrm{k}\right):1.3+2{.3}^{2}+3{.3}^{3}+...+\mathrm{k}.{3}^{\mathrm{k}}=\frac{\left(2\mathrm{k}-1\right){3}^{\mathrm{k}+1}+3}{4}...\left(1\right)\\ \text{We need to prove that}\mathrm{P}\left(\mathrm{k}+1\right)\text{is true whenever P(k) is true.}\\ \text{We have}\\ 1.3+2{.3}^{2}+3{.3}^{3}+...+\mathrm{k}.{3}^{\mathrm{k}}\text{\hspace{0.17em}}+\left(\mathrm{k}+1\right){3}^{\mathrm{k}+1}=\frac{\left\{2\left(\mathrm{k}+1\right)-1\right\}{3}^{\mathrm{k}+2}+3}{4}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\left(2\mathrm{k}-1\right){3}^{\mathrm{k}+1}+3}{4}+\left(\mathrm{k}+1\right){3}^{\mathrm{k}+1}=\frac{\left(2\mathrm{k}+1\right){3}^{\mathrm{k}+2}+3}{4}\text{\hspace{0.17em} …}\left(2\right)\\ \left[\mathrm{From}\text{equation}\left(1\right)\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:\text{\hspace{0.17em}\hspace{0.17em}}\frac{\left(2\mathrm{k}-1\right){3}^{\mathrm{k}+1}+3}{4}+\left(\mathrm{k}+1\right){3}^{\mathrm{k}+1}=\frac{\left(2\mathrm{k}-1\right){3}^{\mathrm{k}+1}+3}{4}+\left(\mathrm{k}+1\right){3}^{\mathrm{k}+1}\\ \text{\hspace{0.17em}}=\frac{\left(2\mathrm{k}-1\right){3}^{\mathrm{k}+1}+3+4\left(\mathrm{k}+1\right){3}^{\mathrm{k}+1}}{4}\\ \text{\hspace{0.17em}}=\frac{\left\{2\mathrm{k}-1+4\left(\mathrm{k}+1\right)\right\}{3}^{\mathrm{k}+1}+3}{4}\\ \text{\hspace{0.17em}}=\frac{\left\{2\mathrm{k}-1+4\mathrm{k}+4\right\}{3}^{\mathrm{k}+1}+3}{4}\\ \text{\hspace{0.17em}}=\frac{\left\{6\mathrm{k}+3\right\}{3}^{\mathrm{k}+1}+3}{4}\\ \text{\hspace{0.17em}}=\frac{\left(2\mathrm{k}+1\right){3}^{\mathrm{k}+2}+3}{4}\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

Q.6 Prove by using principal of mathematical induction for all n ∈ N

$\mathbf{}\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{4}\mathbf{+}\mathbf{\dots }\mathbf{+}\mathbf{n}\mathbf{.}\mathbf{\left(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{\right)}\mathbf{=}\frac{\mathbf{n}\mathbf{\left(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{\right)}\mathbf{\left(}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{\right)}}{\mathbf{3}}\mathbf{.}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{​ be the given statement,}\\ \mathrm{i}.\mathrm{e}.,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{n}\right):\mathrm{}1.2+2.3+3.4+\dots +\mathrm{n}.\left(\mathrm{n}+1\right)=\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)}{3}\\ \mathrm{For}\text{n}=1,\\ \mathrm{P}\left(1\right)=1.2=2\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:\frac{1\left(1+1\right)\left(1+2\right)}{3}=\frac{1\left(2\right)\left(3\right)}{3}\text{\hspace{0.17em}}=2\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k,}\\ \text{i.e., P}\left(\mathrm{k}\right):1.2+2.3+3.4+\dots +\mathrm{k}.\left(\mathrm{k}+1\right)=\frac{\mathrm{k}\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)}{3}...\left(1\right)\\ \text{We need to prove that}\mathrm{P}\left(\mathrm{k}+1\right)\text{is true whenever P(k) is true.}\\ \text{We have}\\ 1.2+2.3+3.4+\dots +\mathrm{k}.\left(\mathrm{k}+1\right)\text{\hspace{0.17em}}+\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)}{3}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{k}\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)}{3}+\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)=\frac{\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)}{3}...\left(2\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{equation}\left(1\right)\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{k}\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)}{3}+\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)=\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\frac{\mathrm{k}}{3}+1\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)}{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

Q.7 Prove by using principal of mathematical induction for all n ∈ N

$\mathbf{1}\mathbf{.}\mathbf{3}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{7}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}\mathbf{\left(}\mathbf{2}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{\right)}\mathbf{\left(}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{\right)}\mathbf{=}\frac{\mathbf{n}\mathbf{\left(}\mathbf{4}{\mathbf{n}}^{\mathbf{2}}\mathbf{+}\mathbf{6}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{\right)}}{\mathbf{3}}\mathbf{.}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{​ be the given statement,}\\ \mathrm{i}.\mathrm{e}.,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{n}\right):\mathrm{}1.3+3.5+5.7+...+\left(2\mathrm{n}-1\right)\left(2\mathrm{n}+1\right)=\frac{\mathrm{n}\left(4{\mathrm{n}}^{2}+6\mathrm{n}-1\right)}{3}\\ \mathrm{For}\text{n}=1,\\ \mathrm{P}\left(1\right)=1.3=3\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1.\left(4×{1}^{2}+6×1-1\right)}{3}=\frac{1\left(9\right)}{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k,}\\ \text{i.e., P}\left(\mathrm{k}\right):1.3+3.5+5.7+...+\left(2\mathrm{k}-1\right)\left(2\mathrm{k}+1\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{k}\left(4{\mathrm{k}}^{2}+6\mathrm{k}-1\right)}{3}...\left(1\right)\\ \text{We need to prove that}\mathrm{P}\left(\mathrm{k}+1\right)\text{is true whenever P(k) is true.}\\ \text{We have}\\ 1.3+3.5+5.7+...+\left(2\mathrm{k}-1\right)\left(2\mathrm{k}+1\right)+\text{\hspace{0.17em}}\left(2\mathrm{k}+1\right)\left(2\mathrm{k}+3\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{k}+1\right)\left\{4{\left(\mathrm{k}+1\right)}^{2}+6\left(\mathrm{k}+1\right)-1\right\}}{3}\\ ⇒\frac{\mathrm{k}\left(4{\mathrm{k}}^{2}+6\mathrm{k}-1\right)}{3}+\left(2\mathrm{k}+1\right)\left(2\mathrm{k}+3\right)=\frac{\left(\mathrm{k}+1\right)\left\{4\left({\mathrm{k}}^{2}+2\mathrm{k}+1\right)+6\mathrm{k}+6-1\right\}}{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{equation}\left(1\right)\right]\\ ⇒\frac{\mathrm{k}\left(4{\mathrm{k}}^{2}+6\mathrm{k}-1\right)}{3}+\left(2\mathrm{k}+1\right)\left(2\mathrm{k}+3\right)=\frac{\left(\mathrm{k}+1\right)\left(4{\mathrm{k}}^{2}+8\mathrm{k}+4+6\mathrm{k}+6-1\right)}{3}\\ ⇒\frac{\mathrm{k}\left(4{\mathrm{k}}^{2}+6\mathrm{k}-1\right)}{3}+\left(2\mathrm{k}+1\right)\left(2\mathrm{k}+3\right)=\frac{\left(\mathrm{k}+1\right)\left(4{\mathrm{k}}^{2}+14\mathrm{k}+9\right)}{3}...\left(2\right)\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}}\frac{\mathrm{k}\left(4{\mathrm{k}}^{2}+6\mathrm{k}-1\right)}{3}+\left(2\mathrm{k}+1\right)\left(2\mathrm{k}+3\right)=\frac{\mathrm{k}\left(4{\mathrm{k}}^{2}+6\mathrm{k}-1\right)}{3}+4{\mathrm{k}}^{2}+6\mathrm{k}+2\mathrm{k}+3\\ =\frac{4{\mathrm{k}}^{3}+6{\mathrm{k}}^{2}-\mathrm{k}+12{\mathrm{k}}^{2}+18\mathrm{k}+6\mathrm{k}+9}{3}\\ =\frac{4{\mathrm{k}}^{3}+18{\mathrm{k}}^{2}+23\mathrm{k}+9}{3}\\ =\frac{\left(\mathrm{k}+1\right)\left(4{\mathrm{k}}^{2}+14\mathrm{k}+9\right)}{3}\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

Q.8 Prove by using principal of mathematical induction for all n ∈ N

$\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{+}\mathbf{2}\mathbf{.}{\mathbf{2}}^{\mathbf{2}}\mathbf{+}\mathbf{3}\mathbf{.}{\mathbf{2}}^{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}\mathbf{n}\mathbf{.}{\mathbf{2}}^{\mathbf{n}}\mathbf{=}\mathbf{\left(}\mathbf{n}\mathbf{–}\mathbf{1}\mathbf{\right)}{\mathbf{2}}^{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{+}\mathbf{2}\mathbf{.}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{​ be the given statement,}\\ \mathrm{i}.\mathrm{e}.,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{n}\right):\mathrm{}1.2+2{.2}^{2}+3{.2}^{2}+...+\mathrm{n}.{2}^{\mathrm{n}}=\left(\mathrm{n}-1\right){2}^{\mathrm{n}+1}+2\\ \mathrm{For}\text{n}=1,\\ \mathrm{P}\left(1\right):\mathrm{L}.\mathrm{H}.\mathrm{S}.=1.2=2\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:\left(1-1\right){2}^{1+1}+2=2\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k,}\\ \text{i.e., P}\left(\mathrm{k}\right):1.2+2{.2}^{2}+3{.2}^{2}+...+\mathrm{k}.{2}^{\mathrm{k}}=\left(\mathrm{k}-1\right){2}^{\mathrm{k}+1}+2...\left(1\right)\\ \text{We need to prove that}\mathrm{P}\left(\mathrm{k}+1\right)\text{is true whenever P(k) is true.}\\ \text{We have}\\ 1.2+2{.2}^{2}+3{.2}^{2}+...+\mathrm{k}.{2}^{\mathrm{k}}+\text{\hspace{0.17em}}\left(\mathrm{k}+1\right).{2}^{\mathrm{k}+1}\text{\hspace{0.17em}}=\left(\mathrm{k}+1-1\right){2}^{\mathrm{k}+1+1}+2\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{k}-1\right){2}^{\mathrm{k}+1}+2+\left(\mathrm{k}+1\right).{2}^{\mathrm{k}+1}=\mathrm{k}.{2}^{\mathrm{k}+2}+2...\left(2\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{equation}\left(1\right)\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{k}-1\right){2}^{\mathrm{k}+1}+2+\left(\mathrm{k}+1\right).{2}^{\mathrm{k}+1}=\left(\mathrm{k}-1+\mathrm{k}+1\right){2}^{\mathrm{k}+1}+2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(2\mathrm{k}\right){2}^{\mathrm{k}+1}+2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{k}.{2}^{\mathrm{k}+2}+2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

Q.9 Prove by using principal of mathematical induction for all n ∈ N

$\frac{\mathbf{1}}{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{8}}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{2}}^{\mathbf{n}}}\mathbf{=}\mathbf{1}\mathbf{–}\frac{\mathbf{1}}{{\mathbf{2}}^{\mathbf{n}}}\mathbf{.}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{​ be the given statement,}\\ \mathrm{i}.\mathrm{e}.,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{n}\right):\mathrm{}\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{{2}^{\mathrm{n}}}=1-\frac{1}{{2}^{\mathrm{n}}}\\ \mathrm{For}\text{n}=1,\\ \mathrm{P}\left(1\right)=\frac{1}{2}\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:1-\frac{1}{{2}^{1}}=\frac{1}{2}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k,}\\ \text{i.e., P}\left(\mathrm{k}\right):\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{{2}^{\mathrm{k}}}=1-\frac{1}{{2}^{\mathrm{k}}}...\left(1\right)\\ \text{We need to prove that}\mathrm{P}\left(\mathrm{k}+1\right)\text{is true whenever P(k) is true.}\\ \text{We have}\\ \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{{2}^{\mathrm{k}}}+\text{\hspace{0.17em}}\frac{1}{{2}^{\mathrm{k}+1}}\text{\hspace{0.17em}}=1-\frac{1}{{2}^{\mathrm{k}+1}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1-\frac{1}{{2}^{\mathrm{k}}}+\frac{1}{{2}^{\mathrm{k}+1}}=1-\frac{1}{{2}^{\mathrm{k}+1}}...\left(2\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{equation}\left(1\right)\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1-\frac{1}{{2}^{\mathrm{k}}}+\frac{1}{{2}^{\mathrm{k}+1}}=1-\left(\frac{2-1}{{2}^{\mathrm{k}+1}}\right)\\ \text{\hspace{0.17em}}=1-\frac{1}{{2}^{\mathrm{k}+1}}\\ \text{\hspace{0.17em}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

Q.10 Prove by using principal of mathematical induction for all n ∈ N

$\frac{\mathbf{1}}{\mathbf{2}\mathbf{.}\mathbf{5}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{5}\mathbf{.}\mathbf{8}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{8}\mathbf{.}\mathbf{11}}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}\frac{\mathbf{1}}{\mathbf{\left(}\mathbf{3}\mathbf{n}\mathbf{–}\mathbf{1}\mathbf{\right)}\mathbf{\left(}\mathbf{3}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{\right)}}\mathbf{=}\frac{\mathbf{n}}{\mathbf{\left(}\mathbf{6}\mathbf{n}\mathbf{+}\mathbf{4}\mathbf{\right)}}\mathbf{.}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{​ be the given statement,}\\ \mathrm{i}.\mathrm{e}.,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{n}\right):\mathrm{}\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(3\mathrm{n}–1\right)\left(3\mathrm{n}+2\right)}=\frac{\mathrm{n}}{\left(6\mathrm{n}+4\right)}\\ \mathrm{For}\text{n}=1,\\ \mathrm{P}\left(1\right):\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{1}{2.5}=\frac{1}{10}\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:\text{\hspace{0.17em}}\frac{1}{\left(6×1+4\right)}=\frac{1}{10}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k,}\\ \text{i.e., P}\left(\mathrm{k}\right):\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(3\mathrm{k}-1\right)\left(3\mathrm{k}+2\right)}=\frac{\mathrm{k}}{\left(6\mathrm{k}+4\right)}...\left(1\right)\\ \text{We need to prove that}\mathrm{P}\left(\mathrm{k}+1\right)\text{is true whenever P(k) is true.}\\ \text{We have}\\ \frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(3\mathrm{k}-1\right)\left(3\mathrm{k}+2\right)}+\text{\hspace{0.17em}}\frac{1}{\left\{3\left(\mathrm{k}+1\right)-1\right\}\left\{3\left(\mathrm{k}+1\right)+2\right\}}\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{k}+1}{\left\{6\left(\mathrm{k}+1\right)+4\right\}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{k}}{\left(6\mathrm{k}+4\right)}+\frac{1}{\left(3\mathrm{k}+2\right)\left(3\mathrm{k}+5\right)}=\frac{\mathrm{k}+1}{\left(6\mathrm{k}+10\right)}...\left(2\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{equation}\left(1\right)\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{k}}{\left(6\mathrm{k}+4\right)}+\frac{1}{\left(3\mathrm{k}+2\right)\left(3\mathrm{k}+5\right)}=\frac{3{\mathrm{k}}^{2}+5\mathrm{k}+2}{2\left(3\mathrm{k}+2\right)\left(3\mathrm{k}+5\right)}\\ \text{\hspace{0.17em}}=\frac{\left(3\mathrm{k}+2\right)\left(\mathrm{k}+1\right)}{2\left(3\mathrm{k}+2\right)\left(3\mathrm{k}+5\right)}\\ \text{\hspace{0.17em}}=\frac{\mathrm{k}+1}{\left(6\mathrm{k}+10\right)}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\text{\hspace{0.17em}}\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

Q.11 Prove by using principal of mathematical induction for all n ∈ N

$\frac{\mathbf{1}}{\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{.}\mathbf{3}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{.}\mathbf{4}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{3}\mathbf{.}\mathbf{4}\mathbf{.}\mathbf{5}}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}\frac{\mathbf{1}}{\mathbf{n}\mathbf{\left(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{\right)}\mathbf{\left(}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{\right)}}\mathbf{=}\frac{\mathbf{n}\mathbf{\left(}\mathbf{n}\mathbf{+}\mathbf{3}\mathbf{\right)}}{\mathbf{4}\mathbf{\left(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{\right)}\mathbf{\left(}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{\right)}}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{​ be the given statement,}\mathrm{i}.\mathrm{e}.,\text{\hspace{0.17em}\hspace{0.17em}}\\ \mathrm{P}\left(\mathrm{n}\right):\mathrm{}\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\mathrm{n}\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)}=\frac{\mathrm{n}\left(\mathrm{n}+3\right)}{4\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)}\\ \mathrm{For}\text{n}=1,\\ \mathrm{P}\left(1\right):\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{1.2.3}=\frac{1}{6}\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:\frac{1\left(1+3\right)}{4\left(1+1\right)\left(1+2\right)}=\frac{4}{4×2×3}=\frac{1}{6}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k, i.e.,}\\ \text{We have}\\ \text{P}\left(\mathrm{k}\right):\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\mathrm{k}\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{k}\left(\mathrm{k}+3\right)}{4\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)}\text{\hspace{0.17em} \hspace{0.17em}}...\left(1\right)\text{\hspace{0.17em}}\\ \text{We need to prove that}\mathrm{P}\left(\mathrm{k}+1\right)\text{is true whenever P(k) is true.}\\ \frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\mathrm{k}\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)}+\frac{1}{\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{k}+1\right)\left(\mathrm{k}+4\right)}{4\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)}\\ ⇒\frac{\mathrm{k}\left(\mathrm{k}+3\right)}{4\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)}+\frac{1}{\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{k}+1\right)\left(\mathrm{k}+4\right)}{4\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)}\text{\hspace{0.17em} \hspace{0.17em}}...\left(2\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{equation}\left(1\right)\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:\text{\hspace{0.17em}}\frac{\mathrm{k}\left(\mathrm{k}+3\right)}{4\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)}+\frac{1}{\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{k}{\left(\mathrm{k}+3\right)}^{2}+4}{4\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{k}\left({\mathrm{k}}^{2}+6\mathrm{k}+9\right)+4}{4\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{k}}^{3}+6{\mathrm{k}}^{2}+9\mathrm{k}+4}{4\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{k}+1\right)\left({\mathrm{k}}^{2}+5\mathrm{k}+4\right)}{4\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{k}+1\right)\left(\mathrm{k}+1\right)\left(\mathrm{k}+4\right)}{4\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{k}+1\right)\left(\mathrm{k}+4\right)}{4\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

Q.12 Prove by using principal of mathematical induction for all n ∈ N

$\mathbf{a}\mathbf{+}\mathbf{ar}\mathbf{+}{\mathbf{ar}}^{\mathbf{2}}\mathbf{+}{\mathbf{ar}}^{\mathbf{3}}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}{\mathbf{ar}}^{\mathbf{n}\mathbf{–}\mathbf{1}}\mathbf{=}\frac{\mathbf{a}\mathbf{\left(}{\mathbf{r}}^{\mathbf{n}}\mathbf{-}\mathbf{1}\mathbf{\right)}}{\mathbf{r}\mathbf{-}\mathbf{1}}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{​ be the given statement,}\mathrm{i}.\mathrm{e}.,\text{\hspace{0.17em}\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{n}\right):\mathrm{}\mathrm{a}+\mathrm{ar}+{\mathrm{ar}}^{2}+{\mathrm{ar}}^{3}+...+{\mathrm{ar}}^{\mathrm{n}–1}=\frac{\mathrm{a}\left({\mathrm{r}}^{\mathrm{n}}-1\right)}{\mathrm{r}-1}\\ \mathrm{For}\text{n}=1,\\ \mathrm{P}\left(1\right):\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{a}\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{a}\left({\mathrm{r}}^{1}-1\right)}{\mathrm{r}-1}=\mathrm{a}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k,i.e.,}\\ \text{P}\left(\mathrm{k}\right):\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}+\mathrm{ar}+{\mathrm{ar}}^{2}+{\mathrm{ar}}^{3}+...+{\mathrm{ar}}^{\mathrm{k}–1}=\frac{\mathrm{a}\left({\mathrm{r}}^{\mathrm{k}}-1\right)}{\mathrm{r}-1}\text{\hspace{0.17em}}...\left(1\right)\\ \text{We need to prove that}\mathrm{P}\left(\mathrm{k}+1\right)\text{is true whenever P(k) is true.}\\ \text{We have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}+\mathrm{ar}+{\mathrm{ar}}^{2}+{\mathrm{ar}}^{3}+...+{\mathrm{ar}}^{\mathrm{k}–1}+{\mathrm{ar}}^{\mathrm{k}}=\frac{\mathrm{a}\left({\mathrm{r}}^{\mathrm{k}+1}-1\right)}{\mathrm{r}-1}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{a}\left({\mathrm{r}}^{\mathrm{k}}-1\right)}{\mathrm{r}-1}+{\mathrm{ar}}^{\mathrm{k}}=\frac{\mathrm{a}\left({\mathrm{r}}^{\mathrm{k}+1}-1\right)}{\mathrm{r}-1}\text{\hspace{0.17em}\hspace{0.17em}}...\left(2\right)\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{equation}\left(1\right)\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{a}\left({\mathrm{r}}^{\mathrm{k}}-1\right)}{\mathrm{r}-1}+{\mathrm{ar}}^{\mathrm{k}}=\frac{\mathrm{a}\left({\mathrm{r}}^{\mathrm{k}}-1\right)+{\mathrm{ar}}^{\mathrm{k}}\left(\mathrm{r}-1\right)}{\mathrm{r}-1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{ar}}^{\mathrm{k}}-\mathrm{a}+{\mathrm{ar}}^{\mathrm{k}+1}-{\mathrm{ar}}^{\mathrm{k}}}{\mathrm{r}-1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{a}\left({\mathrm{r}}^{\mathrm{k}+1}-1\right)}{\mathrm{r}-1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

Q.13 Prove by using principal of mathematical induction for all n ∈ N

$\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)...\left(1+\frac{\left(2\mathrm{n}+1\right)}{{\mathrm{n}}^{2}}\right)={\left(\mathrm{n}+1\right)}^{2}.$

Ans.

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{​ be the given statement,}\mathrm{i}.\mathrm{e}.,\text{\hspace{0.17em}\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{n}\right):\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)...\left\{1+\frac{\left(2\mathrm{n}+1\right)}{{\mathrm{n}}^{2}}\right\}={\left(\mathrm{n}+1\right)}^{2}\\ \mathrm{For}\text{n}=1,\\ \mathrm{P}\left(1\right)=\text{\hspace{0.17em}\hspace{0.17em}}\left(1+\frac{3}{1}\right)=4\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:{\left(1+1\right)}^{2}=4\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k,i.e.,}\\ \text{P}\left(\mathrm{k}\right):\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)...\left\{1+\frac{\left(2\mathrm{k}+1\right)}{{\mathrm{k}}^{2}}\right\}={\left(\mathrm{k}+1\right)}^{2}\text{\hspace{0.17em} …}\left(1\right)\\ \text{We need to prove that}\mathrm{P}\left(\mathrm{k}+1\right)\text{is true whenever P(k) is true.}\\ \text{We have}\\ \left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)...\left\{1+\frac{\left(2\mathrm{k}+1\right)}{{\mathrm{k}}^{2}}\right\}\left\{1+\frac{\left(2\mathrm{k}+3\right)}{{\left(\mathrm{k}+1\right)}^{2}}\right\}={\left(\mathrm{k}+2\right)}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(\mathrm{k}+1\right)}^{2}\left\{1+\frac{\left(2\mathrm{k}+3\right)}{{\left(\mathrm{k}+1\right)}^{2}}\right\}={\left(\mathrm{k}+2\right)}^{2}\text{\hspace{0.17em} …}\left(2\right)\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{equation}\left(1\right)\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:\text{\hspace{0.17em}}{\left(\mathrm{k}+1\right)}^{2}\left\{1+\frac{\left(2\mathrm{k}+3\right)}{{\left(\mathrm{k}+1\right)}^{2}}\right\}={\left(\mathrm{k}+1\right)}^{2}\left\{\frac{{\left(\mathrm{k}+1\right)}^{2}+\left(2\mathrm{k}+3\right)}{{\left(\mathrm{k}+1\right)}^{2}}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{k}}^{2}+2\mathrm{k}+1+2\mathrm{k}+3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{k}}^{2}+4\mathrm{k}+4\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left(\mathrm{k}+2\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

Q.14 Prove by using principal of mathematical induction for all n ∈ N

$\mathbf{\left(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{1}}\mathbf{\right)}\mathbf{\left(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\right)}\mathbf{\left(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{\right)}\mathbf{.}\mathbf{..}\mathbf{\left(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{n}}\mathbf{\right)}\mathbf{=}\mathbf{\left(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{\right)}\mathbf{.}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{​ be the given statement,}\mathrm{i}.\mathrm{e}.,\text{\hspace{0.17em}\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{n}\right):\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)...\left(1+\frac{1}{\mathrm{n}}\right)=\left(\mathrm{n}+1\right)\\ \mathrm{For}\text{n}=1,\\ \mathrm{P}\left(1\right)=\text{\hspace{0.17em}\hspace{0.17em}}\left(1+\frac{1}{1}\right)=2\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:\left(1+1\right)=2\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k,i.e.,}\\ \text{P}\left(\mathrm{k}\right):\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)...\left(1+\frac{1}{\mathrm{k}}\right)=\left(\mathrm{k}+1\right)...\left(1\right)\\ \text{We need to prove that}\mathrm{P}\left(\mathrm{k}+1\right)\text{is true whenever P(k) is true.}\\ \text{We have}\\ \left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)...\left(1+\frac{1}{\mathrm{k}}\right)\left(1+\frac{1}{\mathrm{k}+1}\right)=\left(\mathrm{k}+2\right)\\ ⇒\left(\mathrm{k}+1\right)\left(1+\frac{1}{\mathrm{k}+1}\right)=\left(\mathrm{k}+2\right)...\left(2\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{equation}\left(\mathrm{i}\right)\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:\left(\mathrm{k}+1\right)\left(1+\frac{1}{\mathrm{k}+1}\right)=\left(\mathrm{k}+1\right)\left(\frac{\mathrm{k}+2}{\mathrm{k}+1}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\mathrm{k}+2\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

Q.15 Prove by using principal of mathematical induction for all n ∈ N

${1}^{2}+{3}^{2}+{5}^{2}+...+{\left(2\mathrm{n}–1\right)}^{2}=\frac{\mathrm{n}\left(2\mathrm{n}–1\right)\left(2\mathrm{n}+1\right)}{3}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{​ be the given statement,}\mathrm{i}.\mathrm{e}.,\text{\hspace{0.17em}\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{n}\right):{1}^{2}+{3}^{2}+{5}^{2}+...+{\left(2\mathrm{n}-1\right)}^{2}=\frac{\mathrm{n}\left(2\mathrm{n}-1\right)\left(2\mathrm{n}+1\right)}{3}\\ \mathrm{For}\text{n}=1,\\ \mathrm{P}\left(1\right)=\text{\hspace{0.17em}\hspace{0.17em}}{1}^{2}=1\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:\frac{1\left(2×1-1\right)\left(2×1+1\right)}{3}=1\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k,i.e.,}\\ \text{P}\left(\mathrm{k}\right):{1}^{2}+{3}^{2}+{5}^{2}+...+{\left(2\mathrm{k}-1\right)}^{2}=\frac{\mathrm{k}\left(2\mathrm{k}-1\right)\left(2\mathrm{k}+1\right)}{3}...\left(1\right)\\ \text{We need to prove that}\mathrm{P}\left(\mathrm{k}+1\right)\text{is true whenever P(k) is true.}\\ \text{We have}\\ {1}^{2}+{3}^{2}+{5}^{2}+...+{\left(2\mathrm{k}-1\right)}^{2}+{\left(2\mathrm{k}+1\right)}^{2}=\frac{\left(\mathrm{k}+1\right)\left(2\mathrm{k}+1\right)\left(2\mathrm{k}+3\right)}{3}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{k}\left(2\mathrm{k}-1\right)\left(2\mathrm{k}+1\right)}{3}+{\left(2\mathrm{k}+1\right)}^{2}=\frac{\left(\mathrm{k}+1\right)\left(2\mathrm{k}+1\right)\left(2\mathrm{k}+3\right)}{3}...\left(2\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{equation}\left(1\right)\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{k}\left(2\mathrm{k}-1\right)\left(2\mathrm{k}+1\right)}{3}+{\left(2\mathrm{k}+1\right)}^{2}=\left(2\mathrm{k}+1\right)\left\{\frac{\mathrm{k}\left(2\mathrm{k}-1\right)}{3}+\left(2\mathrm{k}+1\right)\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(2\mathrm{k}+1\right)\left\{\frac{2{\mathrm{k}}^{2}-\mathrm{k}+6\mathrm{k}+3}{3}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(2\mathrm{k}+1\right)\left\{\frac{2{\mathrm{k}}^{2}+5\mathrm{k}+3}{3}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(2\mathrm{k}+1\right)\left\{\frac{\left(2\mathrm{k}+3\right)\left(\mathrm{k}+1\right)}{3}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{k}+1\right)\left(2\mathrm{k}+1\right)\left(2\mathrm{k}+3\right)}{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

Q.16 Prove by using principal of mathematical induction for all n ∈ N

$\frac{\mathbf{1}}{\mathbf{1}\mathbf{.}\mathbf{4}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{4}\mathbf{.}\mathbf{7}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{7}\mathbf{.}\mathbf{10}}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}\frac{\mathbf{1}}{\mathbf{\left(}\mathbf{3}\mathbf{n}\mathbf{-}\mathbf{2}\mathbf{\right)}\mathbf{\left(}\mathbf{3}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{\right)}}\mathbf{=}\frac{\mathbf{n}}{\mathbf{\left(}\mathbf{3}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{\right)}}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{​ be the given statement,}\mathrm{i}.\mathrm{e}.,\text{\hspace{0.17em}\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{n}\right):\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{\left(3\mathrm{n}-2\right)\left(3\mathrm{n}+1\right)}=\frac{\mathrm{n}}{\left(3\mathrm{n}+1\right)}\\ \mathrm{For}\text{n}=1,\\ \mathrm{P}\left(1\right)=\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{1.4}=\frac{1}{4}\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:\frac{1}{\left(3×1+1\right)}=\frac{1}{4}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k,i.e.,}\\ \text{P}\left(\mathrm{k}\right):\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{\left(3\mathrm{k}-2\right)\left(3\mathrm{k}+1\right)}=\frac{\mathrm{k}}{\left(3\mathrm{k}+1\right)}\text{\hspace{0.17em} …}\left(1\right)\\ \text{We need to prove that}\mathrm{P}\left(\mathrm{k}+1\right)\text{is true whenever P(k) is true.}\\ \text{We have}\\ \frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{\left(3\mathrm{k}-2\right)\left(3\mathrm{k}+1\right)}+\frac{1}{\left(3\mathrm{k}+1\right)\left(3\mathrm{k}+4\right)}=\frac{\mathrm{k}+1}{\left(3\mathrm{k}+4\right)}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{k}}{\left(3\mathrm{k}+1\right)}+\frac{1}{\left(3\mathrm{k}+1\right)\left(3\mathrm{k}+4\right)}=\frac{\mathrm{k}+1}{\left(3\mathrm{k}+4\right)}...\left(2\right)\\ \left[\mathrm{From}\text{equation}\left(1\right)\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:\frac{\mathrm{k}}{\left(3\mathrm{k}+1\right)}+\frac{1}{\left(3\mathrm{k}+1\right)\left(3\mathrm{k}+4\right)}=\frac{1}{\left(3\mathrm{k}+1\right)}\left(\mathrm{k}+\frac{1}{3\mathrm{k}+4}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{\left(3\mathrm{k}+1\right)}\left(\frac{3{\mathrm{k}}^{2}+4\mathrm{k}+1}{3\mathrm{k}+4}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{\left(3\mathrm{k}+1\right)}\left\{\frac{\left(3\mathrm{k}+1\right)\left(\mathrm{k}+1\right)}{3\mathrm{k}+4}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{k}+1}{\left(3\mathrm{k}+4\right)}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

Q.17 Prove by using principal of mathematical induction for all n ∈ N

$\frac{\mathbf{1}}{\mathbf{3}\mathbf{.}\mathbf{5}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{5}\mathbf{.}\mathbf{7}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{7}\mathbf{.}\mathbf{9}}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}\frac{\mathbf{1}}{\mathbf{\left(}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{\right)}\mathbf{\left(}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{3}\mathbf{\right)}}\mathbf{=}\frac{\mathbf{n}}{\mathbf{3}\mathbf{\left(}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{3}\mathbf{\right)}}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{​ be the given statement,}\mathrm{i}.\mathrm{e}.,\text{\hspace{0.17em}\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{n}\right):\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2\mathrm{n}+1\right)\left(2\mathrm{n}+3\right)}=\frac{\mathrm{n}}{3\left(2\mathrm{n}+3\right)}\\ \mathrm{For}\text{n}=1,\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(1\right):\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{3.5}=\frac{1}{15}\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:\text{\hspace{0.17em}}\frac{1}{3\left(2×1+3\right)}\text{\hspace{0.17em}\hspace{0.17em}}=\frac{1}{15}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k,i.e.,}\\ \text{P}\left(\mathrm{k}\right):\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2\mathrm{k}+1\right)\left(2\mathrm{k}+3\right)}=\frac{\mathrm{k}}{3\left(2\mathrm{k}+3\right)}\text{\hspace{0.17em} …}\left(1\right)\\ \text{We need to prove that}\mathrm{P}\left(\mathrm{k}+1\right)\text{is true whenever P(k) is true.}\\ \text{We have}\\ \frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2\mathrm{k}+1\right)\left(2\mathrm{k}+3\right)}+\frac{1}{\left(2\mathrm{k}+3\right)\left(2\mathrm{k}+5\right)}=\frac{\mathrm{k}+1}{3\left(2\mathrm{k}+5\right)}\\ ⇒\frac{\mathrm{k}}{3\left(2\mathrm{k}+3\right)}+\frac{1}{\left(2\mathrm{k}+3\right)\left(2\mathrm{k}+5\right)}=\frac{\mathrm{k}+1}{3\left(2\mathrm{k}+5\right)}\text{\hspace{0.17em} \hspace{0.17em}…}\left(2\right)\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}}\left[\mathrm{From}\text{equation}\left(\mathrm{i}\right)\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:\frac{\mathrm{k}}{3\left(2\mathrm{k}+3\right)}+\frac{1}{\left(2\mathrm{k}+3\right)\left(2\mathrm{k}+5\right)}=\frac{1}{\left(2\mathrm{k}+3\right)}\left\{\frac{\mathrm{k}}{3}+\frac{1}{2\mathrm{k}+5}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{\left(2\mathrm{k}+3\right)}\left\{\frac{2{\mathrm{k}}^{2}+5\mathrm{k}+3}{3\left(2\mathrm{k}+5\right)}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{\left(2\mathrm{k}+3\right)}\left\{\frac{\left(\mathrm{k}+1\right)\left(2\mathrm{k}+3\right)}{3\left(2\mathrm{k}+5\right)}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\mathrm{k}+1\right)}{3\left(2\mathrm{k}+5\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

Q.18 Prove by using principal of mathematical induction for all n ∈ N

$1+2+3+...+\mathrm{n}<\frac{1}{8}{\left(2\mathrm{n}+1\right)}^{2}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{be the given statement, i.e.,}\\ \mathrm{P}\left(\mathrm{n}\right):1+2+3+...+\mathrm{n}<\frac{1}{8}{\left(2\mathrm{n}+1\right)}^{2}\\ \mathrm{P}\left(\mathrm{n}\right)\text{is true when n}=\text{1,}\\ \text{since\hspace{0.17em}1<}\frac{1}{8}{\left(2×1+1\right)}^{2}\\ ⇒1<\frac{9}{8}\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{is true for n}=\text{1.}\\ \text{Let P}\left(\mathrm{n}\right)\text{is true for some natural number k, i.e.,}\\ \mathrm{P}\left(\mathrm{k}\right):\text{\hspace{0.17em}}1+2+3+...+\mathrm{k}<\frac{1}{8}{\left(2\mathrm{k}+1\right)}^{2}...\left(1\right)\\ \text{We need to prove that}\mathrm{P}\left(\mathrm{k}+1\right)\text{is true whenever P(k) is true.}\\ \text{Adding}\left(\mathrm{k}+1\right)\text{both sides of relation}\left(1\right),\text{we get}\\ 1+2+3+...+\mathrm{k}+\left(\mathrm{k}+1\right)<\frac{1}{8}{\left(2\mathrm{k}+1\right)}^{2}+\left(\mathrm{k}+1\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1+2+3+...+\mathrm{k}+\left(\mathrm{k}+1\right)<\frac{1}{8}\left\{\left(4{\mathrm{k}}^{2}+4\mathrm{k}+1\right)+8\left(\mathrm{k}+1\right)\right\}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1+2+3+...+\mathrm{k}+\left(\mathrm{k}+1\right)<\frac{1}{8}\left(4{\mathrm{k}}^{2}+12\mathrm{k}+9\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1+2+3+...+\mathrm{k}+\left(\mathrm{k}+1\right)<\frac{1}{8}{\left(2\mathrm{k}+3\right)}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1+2+3+...+\mathrm{k}+\left(\mathrm{k}+1\right)<\frac{1}{8}{\left\{2\left(\mathrm{k}+1\right)+1\right\}}^{2}\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

Q.19 Prove by using principal of mathematical induction for all n ∈ N n (n + 1) (n + 5) is a multiple of 3.

Ans.

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{be the given statement, i.e.,}\\ \mathrm{P}\left(\mathrm{n}\right):\text{n}\left(\text{n}+\text{1}\right)\text{}\left(\text{n}+\text{5}\right)\text{is a multiple of 3}\\ \mathrm{Putting}\text{n}=\text{1}\\ \mathrm{P}\left(1\right):\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}1}\left(\text{1}+\text{1}\right)\left(\text{1}+\text{5}\right)=1.2.6\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=12,\text{which is a multiple of 3.}\\ \therefore \mathrm{P}\left(\mathrm{n}\right)\text{is true when n}=\text{1.}\\ \text{Let P}\left(\mathrm{n}\right)\text{is true for some natural number k, i.e.,}\\ \mathrm{P}\left(\mathrm{k}\right):\text{k}\left(\text{k}+\text{1}\right)\left(\text{k}+\text{5}\right)\text{is divisible by 3}\\ \mathrm{P}\left(\mathrm{k}\right):\text{\hspace{0.17em}\hspace{0.17em}k}\left(\text{k}+\text{1}\right)\left(\text{k}+\text{5}\right)=3\mathrm{\lambda },\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{where}\text{\hspace{0.17em}}\mathrm{\lambda }\in \mathrm{N}...\left(1\right)\\ \text{We need to prove that}\mathrm{P}\left(\mathrm{k}+1\right)\text{is true whenever P(k) is true.}\\ \therefore \left(\text{k+1}\right)\left\{\left(\text{k}+\text{1}\right)+1\right\}\left\{\left(\text{k}+\text{5}\right)+1\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\text{k+1}\right)\left(\text{k}+\text{2}\right)\left\{\left(\text{k}+\text{5}\right)+1\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\text{k+1}\right)\left(\text{k}+\text{2}\right)\left(\text{k}+\text{5}\right)+\left(\text{k+1}\right)\left(\text{k}+\text{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{k}\left(\text{k}+\text{1}\right)\left(\text{k}+\text{5}\right)+2\left(\text{k}+\text{1}\right)\left(\text{k}+\text{5}\right)+\left(\text{k+1}\right)\left(\text{k}+\text{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\mathrm{\lambda }+\left(\text{k}+\text{1}\right)\left\{2\left(\text{k}+\text{5}\right)+\left(\text{k}+\text{2}\right)\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\mathrm{\lambda }+\left(\text{k}+\text{1}\right)\left(\text{2k}+\text{10}+\text{k}+\text{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\mathrm{\lambda }+\left(\text{k}+\text{1}\right)\left(3\mathrm{k}+12\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\mathrm{\lambda }+3\left(\text{k}+\text{1}\right)\left(\mathrm{k}+4\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\left\{\mathrm{\lambda }+\left(\text{k}+\text{1}\right)\left(\mathrm{k}+4\right)\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{k}×\mathrm{p},\mathrm{where}\text{\hspace{0.17em}}\mathrm{p}=\left\{\mathrm{\lambda }+\left(\text{k}+\text{1}\right)\left(\mathrm{k}+4\right)\right\}\in \mathrm{N}\\ \mathrm{Therefore},\text{​ P}\left(\mathrm{k}+1\right)\text{is a multiple of 3.}\\ \text{Thus, P}\left(\mathrm{k}+1\right)\text{is true whenever P}\left(\mathrm{k}\right)\text{is true.}\\ \text{Hence, by principle of mathematical induction, P(n) is true for}\\ \text{all n}\in \text{N.}\end{array}$

Q.20 Prove by using principal of mathematical induction for all n ∈ N 102n – 1+ 1 is divisible by 11.

Ans.

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{be the given statement,}\\ \text{i.e.,}\mathrm{P}\left(\mathrm{n}\right):\text{1}{0}^{\text{2n}–\text{1}}+1\text{is divisible by 11}\\ \mathrm{Putting}\text{n}=\text{1}\\ \mathrm{P}\left(1\right):\text{1}{0}^{\text{2}\left(1\right)\text{}–\text{1}}+1=10+1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=11,\text{which is a multiple of 11.}\\ \therefore \mathrm{P}\left(\mathrm{n}\right)\text{is true when n}=\text{1.}\\ \text{Let P}\left(\mathrm{n}\right)\text{is true for some natural number k, i.e.,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{k}\right):\text{1}{0}^{\text{2k}–\text{1}}+1\text{\hspace{0.17em}is divisible by 11}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{k}\right):\text{\hspace{0.17em}\hspace{0.17em}1}{0}^{\text{2k}–\text{1}}+1=11\mathrm{\lambda },\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{where}\text{\hspace{0.17em}}\mathrm{\lambda }\in \mathrm{N}...\left(1\right)\\ \text{We need to prove that}\mathrm{P}\left(\mathrm{k}+1\right)\text{is true whenever P(k) is true.}\\ \therefore \text{1}{0}^{\text{2}\left(\text{k+1}\right)\text{}–\text{1}}+1=\text{1}{0}^{\text{2k+2}–\text{1}}+1\\ \text{\hspace{0.17em}\hspace{0.17em}}=\text{1}{0}^{\text{2k+1}}+1\\ \text{\hspace{0.17em}\hspace{0.17em}}={10}^{2}\left(\text{1}{0}^{\text{2k-1}}+1-1\right)+1\\ \text{\hspace{0.17em}\hspace{0.17em}}={10}^{2}\left(11\mathrm{\lambda }-1\right)+1\left[\mathrm{From}\text{equation}\left(1\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}=1100\mathrm{\lambda }-100+1\\ \text{\hspace{0.17em}\hspace{0.17em}}=1100\mathrm{\lambda }-99\\ \text{\hspace{0.17em}\hspace{0.17em}}=11\left(100\mathrm{\lambda }-9\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=11×\mathrm{p},\text{where \hspace{0.17em}p}=\left(100\mathrm{\lambda }-9\right)\in \mathrm{N}\\ \mathrm{Therefore},\text{​ P}\left(\mathrm{k}+1\right)\text{is a multiple of 11.}\\ \text{Thus, P}\left(\mathrm{k}+1\right)\text{is true whenever P}\left(\mathrm{k}\right)\text{is true.}\\ \text{Hence, by principle of mathematical induction, P(n) is true for}\\ \text{all n}\in \text{N.}\end{array}$

Q.21 Prove by using principal of mathematical induction for all n ∈ N x2n – y2n is divisible by x + y.

Ans.

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{be the given statement, i.e.,}\\ \mathrm{P}\left(\mathrm{n}\right):{\text{x}}^{\text{2n}}-{\text{y}}^{\text{2n}}\text{is divisible by x}+\text{y}\\ \mathrm{Putting}\text{n}=\text{1}\\ \mathrm{P}\left(1\right):{\text{x}}^{2\left(1\right)}-{\text{y}}^{\text{2}\left(1\right)}={\mathrm{x}}^{2}-{\mathrm{y}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\mathrm{x}-\mathrm{y}\right)\left(\mathrm{x}+\mathrm{y}\right),\text{which is divisible by x}+\text{y.}\\ \therefore \mathrm{P}\left(\mathrm{n}\right)\text{is true when n}=\text{1.}\\ \text{Let P}\left(\mathrm{n}\right)\text{is true for some natural number k, i.e.,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{k}\right):{\text{x}}^{\text{2k}}-{\text{y}}^{\text{2k}}\text{\hspace{0.17em}\hspace{0.17em}is divisible by x}+\text{y}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{k}\right):{\text{\hspace{0.17em}\hspace{0.17em}x}}^{\text{2k}}-{\text{y}}^{\text{2k}}=\left(\text{x}+\text{y}\right)\mathrm{\lambda },\text{\hspace{0.17em}}\mathrm{where}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{\lambda }\in \mathrm{N}...\left(1\right)\\ \text{We need to prove that}\mathrm{P}\left(\mathrm{k}+1\right)\text{is true whenever P(k) is true.}\\ \therefore {\text{x}}^{\text{2}\left(\text{k+1}\right)}-{\text{y}}^{\text{2}\left(\text{k+1}\right)}={\text{x}}^{\text{2}\left(\text{k+1}\right)}-{\text{y}}^{\text{2}\left(\text{k+1}\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{x}}^{\text{2k+2}}-{\mathrm{y}}^{\text{2k+2}}\\ \text{\hspace{0.17em}\hspace{0.17em}}={\mathrm{x}}^{2}.{\mathrm{x}}^{2\mathrm{k}}-{\mathrm{x}}^{2}.{\mathrm{y}}^{2\mathrm{k}}+{\mathrm{x}}^{2}.{\mathrm{y}}^{2\mathrm{k}}-{\mathrm{y}}^{2}.{\mathrm{y}}^{2\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}}={\mathrm{x}}^{2}\left({\mathrm{x}}^{2\mathrm{k}}-{\mathrm{y}}^{2\mathrm{k}}\right)+{\mathrm{y}}^{2\mathrm{k}}\left({\mathrm{x}}^{2}-{\mathrm{y}}^{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}={\mathrm{x}}^{2}\left(\text{x}+\text{y}\right)\mathrm{\lambda }+{\mathrm{y}}^{2\mathrm{k}}\left(\mathrm{x}-\mathrm{y}\right)\left(\mathrm{x}+\mathrm{y}\right)\left[\mathrm{From}\text{equation}\left(1\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}=\left(\mathrm{x}+\mathrm{y}\right)\left\{{\mathrm{x}}^{2}\mathrm{\lambda }+{\mathrm{y}}^{2\mathrm{k}}\left(\mathrm{x}-\mathrm{y}\right)\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\mathrm{which}\text{\hspace{0.17em}is a factor of}\left(\mathrm{x}+\mathrm{y}\right)\\ \mathrm{Therefore},\text{​ P}\left(\mathrm{k}+1\right)\text{is divisible by}\left(\mathrm{x}+\mathrm{y}\right)\text{.}\\ \text{Thus, P}\left(\mathrm{k}+1\right)\text{is true whenever P}\left(\mathrm{k}\right)\text{is true.}\\ \text{Hence, by principle of mathematical induction, P(n) is true for}\\ \text{all n}\in \text{N.}\end{array}$

Q.22 Prove by using principal of mathematical induction for all n ∈ N 32n+2 – 8n – 9 is divisible by 8.

Ans.

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{be the given statement, i.e.,}\\ \mathrm{P}\left(\mathrm{n}\right):{\text{3}}^{\text{2n}+\text{2}}-\text{8n}-\text{9 is divisible by 8}\\ \mathrm{Putting}\text{n}=\text{1}\\ \mathrm{P}\left(1\right):{\text{3}}^{\text{2}\left(1\right)+\text{2}}-\text{8}\left(1\right)-\text{9}={\text{3}}^{4}-\text{8}-\text{9}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=64,\text{which is divisible by 8}\\ \therefore \mathrm{P}\left(\mathrm{n}\right)\text{is true when n}=\text{1.}\\ \text{Let P}\left(\mathrm{n}\right)\text{is true for some natural number k, i.e.,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{k}\right):{\text{3}}^{\text{2k}+\text{2}}-\text{8k}-\text{9 is divisible by 8}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{k}\right):{\text{\hspace{0.17em}\hspace{0.17em}3}}^{\text{2k}+\text{2}}-\text{8k}-\text{9}=8\mathrm{\lambda },\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{where}\text{\hspace{0.17em}}\mathrm{\lambda }\in \mathrm{N}...\left(1\right)\\ \text{We need to prove that}\mathrm{P}\left(\mathrm{k}+1\right)\text{is true whenever P(k) is true.}\\ \therefore {\text{3}}^{\text{2}\left(\text{k+1}\right)+\text{2}}-\text{8}\left(\text{k+1}\right)-\text{9}={\text{3}}^{\text{2k+4}}-\text{8k}-\text{8}-\text{9}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={3}^{2}{\text{.3}}^{\text{2k+2}}-\text{8k}-17\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=9\left(8\mathrm{\lambda }+8\mathrm{k}+9\right)-\text{8k}-17\text{\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{equation}\left(1\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=72\mathrm{\lambda }+72\mathrm{k}+81-\text{8k}-17\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=72\mathrm{\lambda }+64\mathrm{k}+64\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8\left(9\mathrm{\lambda }+8\mathrm{k}+8\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8×\mathrm{p},\text{where \hspace{0.17em}p}=\left(9\mathrm{\lambda }+8\mathrm{k}+8\right)\in \mathrm{N}\\ \mathrm{Therefore},\text{​ P}\left(\mathrm{k}+1\right)\text{is divisible by 8.}\\ \text{Thus, P}\left(\mathrm{k}+1\right)\text{is true whenever P}\left(\mathrm{k}\right)\text{is true.}\\ \text{Hence, by principle of mathematical induction, P(n) is true for}\\ \text{all n}\in \text{N.}\end{array}$

Q.23 Prove by using principal of mathematical induction for all n ∈ N 41n – 14n is multiple of 27.

Ans.

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{be the given statement, i.e.,}\\ \mathrm{P}\left(\mathrm{n}\right):{\left(\text{41}\right)}^{\text{n}}-{\left(\text{14}\right)}^{\text{n}}\text{\hspace{0.17em}is multiple of 27}\\ \mathrm{Putting}\text{n}=\text{1}\\ \mathrm{P}\left(1\right):{\left(\text{41}\right)}^{\text{1}}-{\left(\text{14}\right)}^{\text{1}}=41-14\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=27,\text{which is divisible by 27}\\ \therefore \mathrm{P}\left(\mathrm{n}\right)\text{is true when n}=\text{1.}\\ \text{Let P}\left(\mathrm{n}\right)\text{is true for some natural number k, i.e.,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{k}\right):{\left(\text{41}\right)}^{\text{k}}-{\left(\text{14}\right)}^{\text{k}}\text{\hspace{0.17em}is multiple of 27}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\left(\mathrm{k}\right):\text{\hspace{0.17em}}{\left(\text{41}\right)}^{\text{k}}-{\left(\text{14}\right)}^{\text{k}}=27\mathrm{\lambda },\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{where}\text{\hspace{0.17em}}\mathrm{\lambda }\in \mathrm{N}...\left(1\right)\\ \text{We need to prove that}\mathrm{P}\left(\mathrm{k}+1\right)\text{is true whenever P(k) is true.}\\ \therefore {\left(\text{41}\right)}^{\text{k+1}}-{\left(\text{14}\right)}^{\text{k+1}}=\text{41}{\left(41\right)}^{\mathrm{k}}-14{\left(14\right)}^{\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=41\left\{27\mathrm{\lambda }+{\left(14\right)}^{\mathrm{k}}\right\}-14{\left(14\right)}^{\mathrm{k}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{equation}\left(1\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=41×27\mathrm{\lambda }+41×{\left(14\right)}^{\mathrm{k}}-14{\left(14\right)}^{\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=41×27\mathrm{\lambda }+\left(41-14\right){\left(14\right)}^{\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=41×27\mathrm{\lambda }+27{\left(14\right)}^{\mathrm{k}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=27\left\{41\text{\hspace{0.17em}}\mathrm{\lambda }+{\left(14\right)}^{\mathrm{k}}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=27×\mathrm{p},\text{where \hspace{0.17em}p}=\left\{41\text{\hspace{0.17em}}\mathrm{\lambda }+{\left(14\right)}^{\mathrm{k}}\right\}\in \mathrm{N}\\ \mathrm{Therefore},\text{​ P}\left(\mathrm{k}+1\right)\text{is divisible by 27.}\\ \text{Thus, P}\left(\mathrm{k}+1\right)\text{is true whenever P}\left(\mathrm{k}\right)\text{is true.}\\ \text{Hence, by principle of mathematical induction, P(n) is true for}\\ \text{all n}\in \text{N.}\end{array}$

Q.24 Prove by using principal of mathematical induction for all n ∈ N (2n + 7) < (n + 3)2.

Ans.

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{be the given statement, i.e.,}\\ \mathrm{P}\left(\mathrm{n}\right):\left(\text{2n}+\text{7}\right)<{\left(\text{n}+\text{3}\right)}^{\text{2}}\\ \mathrm{P}\left(\mathrm{n}\right)\text{is true when n}=\text{1,}\\ \left(\text{2}×\text{1}+\text{7}\right)<{\left(\text{1}+\text{3}\right)}^{\text{2}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}9<16\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{is true for n}=\text{1.}\\ \text{Let P}\left(\mathrm{n}\right)\text{is true for some natural number k, i.e.,}\\ \mathrm{P}\left(\mathrm{k}\right):\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\text{2k}+\text{7}\right)<{\left(\text{k}+\text{3}\right)}^{\text{2}}...\left(1\right)\\ \text{We need to prove that}\mathrm{P}\left(\mathrm{k}+1\right)\text{is true whenever P(k) is true.}\\ \mathrm{So},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left\{\text{2}\left(\text{k}+\text{1}\right)+\text{7}\right\}<{\left\{\left(\text{k}+\text{1}\right)+\text{3}\right\}}^{\text{2}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\text{2k}+\text{9}\right)<{\left(\text{k}+\text{4}\right)}^{2}...\left(2\right)\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:\left\{\text{2}\left(\text{k}+\text{1}\right)+\text{7}\right\}=\left(2\mathrm{k}+7\right)+2\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left\{\text{2}\left(\text{k}+\text{1}\right)+\text{7}\right\}<{\left(\text{k}+\text{3}\right)}^{\text{2}}+2\left[\mathrm{From}\text{}\left(1\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left\{\text{2}\left(\text{k}+\text{1}\right)+\text{7}\right\}<{\mathrm{k}}^{2}+6\mathrm{k}+9+2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left\{\text{2}\left(\text{k}+\text{1}\right)+\text{7}\right\}<{\mathrm{k}}^{2}+6\mathrm{k}+11...\left(3\right)\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{k}}^{2}+6\mathrm{k}+11<{\mathrm{k}}^{2}+8\mathrm{k}+16\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{k}}^{2}+6\mathrm{k}+11<{\left(\mathrm{k}+4\right)}^{2}...\left(4\right)\\ \mathrm{From}\text{relation}\left(3\right)\text{and relation}\left(4\right),\text{​ we have}\\ \left\{\text{2}\left(\text{k}+\text{1}\right)+\text{7}\right\}<{\left(\mathrm{k}+4\right)}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left\{\text{2}\left(\text{k}+\text{1}\right)+\text{7}\right\}<{\left\{\left(\mathrm{k}+1\right)+3\right\}}^{2}\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

## Please register to view this section

### 1. Where are the concepts to Derive the Sum of Natural Numbers present for practice?

The Concept to Derive the Sum of Natural Numbers are present at the NCERT Solutions for Class 11 Maths Chapter 4 Exercise 4.1. The Extramarks’ website provides access to the NCERT Solutions for Class 11 Maths Chapter 4 Exercise 4.1. These can be downloaded or accessed online.

### 2. What is the meaning of Sets, Relation and Functions?

Sets in Mathematics means a list of Objects or Elements which are of definitive value. This concept has been discussed in detail in the NCERT Solutions for Class 11 Maths Chapter 4 Exercise 4.1.

### 3. How is the Chain of Natural Numbers calculated?

The chain of Natural Numbers is created with the help of the concept of the Principle of Mathematical Induction where the variable n is taken as 1. The concept is already present at the NCERT Solutions for Class 11 Maths Chapter 4 Exercise 4.1 which can be accessed on the Extramarks’ website.

### 4. Where to find the Principle of Mathematical Induction and why should it be studied?

The Principle of Mathematical Induction can be found in the NCERT Solutions for Class 11 Maths Chapter 4 Exercise 4.1 which can be accessed through the Extramarks website. It is important to study the topic in detail and practice it to master it if a student aspires to study Engineering or specialise in studies with Mathematics. The NCERT Solutions for Class 11 Maths Chapter 4 Exercise 4.1 help in boosting the students overall preparation for the final examination.

### 5. Where to find help to understand Quadratic Equations and Complex numbers?

Complex numbers and Quadratic Equations has been covered in the NCERT Solutions for Class 11 Maths Chapter 4 Exercise 4.1 which is available on the Extramarks website.

The frequently asked questions are a part of a chapter or a topic which focuses on doubts and misunderstandings of students which is common. It is not accumulated on the basics of a batch of students from a single year but these questions are repetitive among students through several years. This makes the questions genuine and the answers give a chance to the students to get their enquiries answered quickly and efficiently. The NCERT Solutions for Class 11 Maths Chapter 4 Exercise 4.1 provides an example of the assistance provided to the students.

As mentioned that the NCERT Solutions For Class 11 Maths Chapter 4 Exercise 4.1  is available on the Extramarks’ website, there are many other Chapters of many other subjects which are available on this platform. The Extramarks’ website is a very useful tool which enhances the chances of the students to understand a subject better. As mentioned, it also has several PDF files which can be downloaded and can be accessed accordingly to assist the students. The NCERT Solutions for Class 11 Maths Chapter 4 Exercise 4.1 which is accessible on the Extramarks’ website shows the level of proficiency with which it helps the students.