# NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem (Ex 8.2) Exercise 8.2

Mathematics is one of the most important practical and conceptual subjects. The applications of mathematics can be found in theories of Physics, Chemistry, Biology, Statistics, etc. It is used by everyone every day in some form or another. Various concepts of Mathematics are applied in businesses, in engineering, and in day-to-day things like using a calculator. Therefore, Mathematics becomes a very important subject and is taught to children from a very young age. A few tips that students should keep in mind while studying Class 11 Mathematics are:

1. Students should begin by understanding the derivations of the formulas and theorems. This will help them understand the chapter and its origin.
2. They should make short notes of all the important formulas to revise before the exam and also note down the theorems separately so that they can prove the theorems in tests.
3. They should start by solving easy questions first, where direct use of formulas is needed. This will help them get used to the various formulas in the chapter. Solving the easy questions first will also give them confidence.
4. They should then move on to the Higher Order Thinking Skills (HOTS) questions. These questions will help them use their knowledge to the fullest and make them exam ready.

If students face any difficulty in solving any of the NCERT exercise questions or they are looking for the best solutions to the questions, then they can refer to the Extramarks NCERT solutions for Mathematics Class 11. To find the best and easiest-to-understand solutions for Chapter 8 Class 11 Exercise 8.2, students should refer to the  NCERT Solutions for Class 11 Maths Chapter 8 Exercise 8.2. The NCERT Solutions For Class 11 Maths Chapter 8 Exercise 8.2, can be found on the Extramarks website and mobile application. The  NCERT Solutions For Class 11 Maths Chapter 8 Exercise 8.2 can be downloaded in PDF format for offline access.

A lot of students find Class 11 Mathematics difficult. To make studying Class 11 Mathematics easier for them, here are a few tips:

1. Students should attend Mathematics classes regularly. Missing even one lecture can confuse them in the next one. Attending regular lectures ensures continuity in learning the concepts, which are very important.
2. Students can make short notes in class. These short notes should include examples of questions taught during the class. Students can also refer to the NCERT revision notes offered by Extramarks for Class 11 Mathematics. Extramarks provides revision notes for all classes and all subjects. The notes provided by Extramarks are reliable and authentic, and they cover the whole syllabus holistically.
3. After the class, students should always practice a few questions. By doing so, they will be able to retain what was taught in the class. In case of any difficulty, they can refer to the NCERT Solutions for Class 11 Maths Chapter 8 Exercise 8.2. They can also download the NCERT Solutions For Class 11 Maths Chapter 8 Exercise 8.2 from the Extramarks website and mobile application.
4. Students should focus on all the chapters and complete the whole syllabus to prepare well for the exams. Each chapter should be revised thoroughly. Learners can find the NCERT solutions for all Class 11 Mathematics chapters on the Extramarks website and mobile application. They should refer to those solutions for the practice of important questions.
5. Students should regularly practice mock tests and sample papers to be exam ready. They can find the sample papers on the official CBSE website. Self-assessing their preparation will be very beneficial for them.
6. Lastly, Class 11 Mathematics should be studied very seriously as it forms the basis of Class 12 Mathematics, Physics, and also Chemistry.

Mathematics is a very scoreable subject. If students prepare well for this subject, it will become easier for them to study Physics, Chemistry, and Biology. Moreover, studying Mathematics well will help them choose their future careers. To prepare better for the NCERT Solutions for Class 11 Maths Chapter 8 Exercise 8.2, students can refer to the Extramarks website and mobile application. The NCERT Solutions For Class 11 Maths Chapter 8 Exercise 8.2, will benefit them a lot in their preparation for their exams.

## NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem (Ex 8.2) Exercise 8.2

The best way to score well in Mathematics is to practice as many questions as one can. With ample practice, students do not make common mistakes and eventually gain confidence. They should solve questions of all varieties to answer better in the examination hall when appearing for the exam. Chapter 8 in Class 11 Mathematics is Binomial Theorem. To gain confidence in this chapter, students should start by understanding the formula derivations, and theorems and then apply those to the questions. To get the solutions for Chapter 8, students can click the link below and download the NCERT Solutions for Class 11 Maths Chapter 8 Exercise 8.2. For the convenience of students, the NCERT Solutions For Class 11 Maths Chapter 8 Exercise 8.2 can be downloaded in PDF format. They can download the NCERT Solutions from Extramarks for all Class 11 subjects. Moreover, all the solutions are available on the Extramarks website and mobile application.

The NCERT Solutions for all subjects can be found on the Extramarks website and mobile application. The NCERT Solutions Class 12, NCERT Solutions Class 11, NCERT Solutions Class 10, NCERT Solutions Class 9, NCERT Solutions Class 8, NCERT Solutions Class 7, NCERT Solutions Class 6, NCERT Solutions Class 5, NCERT Solutions Class 4, NCERT Solutions Class 3, NCERT Solutions Class 2, NCERT Solutions Class 1 for all subjects can be downloaded from the Extramarks website and mobile application. The NCERT solutions can be downloaded in PDF format. These NCERT solutions will help students learn and understand the subjects better. These solutions provided by Extramarks are trustworthy and written by expert teachers.

### What Is The General and  Middle Term Binomial Expression?

The algebraic expansion of a binomial’s powers is known as the Binomial Theorem or expansion. This theorem states that the polynomial “(a + b)n” may be expanded into a sum involving terms of the form “axzyc,” where the exponents z and c are non-negative integers and z + c = n, and the coefficient of each term is an integer that depends on the values of n and b.

By using the Binomial Theorem Formula, we may determine a binomial’s power without having to multiply it tediously. The formula is also used to assist one to find the general and middle terms in the algebraic expression expansion. Students can use the  NCERT Solutions For Class 11 Maths Chapter 8 Exercise 8.2 to solve Ex 8.2 Class11. They can download the  NCERT Solutions For Class 11 Maths Chapter 8 Exercise 8.2 from the Extramarks website and mobile application.

The topics and concepts included in the chapter named – Binomial Theorem Class 11 are:

• Introduction
• Binomial Theorem for positive integral indices
• Binomial Theorem for any positive integer n
• Special Cases
• General and Middle Term

The Binomial Theorem, the Binomial Theorem for positive integral indices, Pascal’s triangle, the general term and middle terms in the expansion of (a + b)n, the middle term, the Binomial Theorem for any index, the binomial coefficients, the properties of the binomial coefficient, some significant results, and multinomial expansion are all covered in this chapter.

Students should solve all the NCERT questions given at the end of the chapter for practice. They should highlight the difficult questions and practice them again and again. While practising questions, students should ensure that they are not making any calculation mistakes and using the right formulas. They should present their answers in neat handwriting for the convenience of the examiner. Extramarks provides the detailed NCERT Solutions for Class 11 Maths Chapter 8 Exercise 8.2.

Applications of the Binomial Theorem: The Binomial Theorem has several uses. The following applications are only a few:

• The Binomial Theorem is extensively used in the analysis of statistics and probability. Given the current reliance of our economies and enterprises on statistical and probability analysis, it is particularly helpful. Students should take the Binomial Theorem seriously if they intend to pursue an occupation in this field.
• The Binomial Theorem is used in Advanced Mathematics to find the roots of equations with higher powers.
• The Binomial Theorem plays a crucial role in Algebra.
• Additionally, it has applications in the areas of Matrices, Mathematical Induction, Permutations and Combinations, and so forth.

To solve questions from the above topics, students should refer to Class 11 Maths Chapter 8 Exercise 8.2 solutions. They can find the NCERT Solutions For Class 11 Maths Chapter 8 Exercise 8.2 on the Extramarks website and mobile application. The NCERT Solutions For Class 11 Maths Chapter 8 Exercise 8.2 are very reliable and authentic, curated by expert Mathematics teachers.

### Access NCERT Solutions for Class 11 Maths Chapter 8 – Binomial Theorem

Students can download the NCERT Solutions For Class 11 Maths Chapter 8 Exercise 8.2, from the Extramarks website and mobile application. They can use the NCERT Solutions For Class 11 Maths Chapter 8 Exercise 8.2 to clear their doubts and find solutions to the difficult questions of Exercise 8.2.

They can download the NCERT Solutions for Class 11 Maths Chapter 8 Exercise 8.1, in PDF format to be able to study offline too. While studying the NCERT Solutions for Class 11 Maths Chapter 8 Exercise 8.1, students should focus on understanding the important formulas and concepts. They should practice as many questions as they can.

### NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Exercise 8.2

The NCERT Solutions For Class 11 Maths Chapter 8 Exercise 8.2 will benefit students in understanding chapter 8 of Class 11 better. The NCERT Solutions For Class 11 Maths Chapter 8 Exercise 8.2 can be used to learn how to write the solutions properly in the examination. NCERT solutions in NCERT Solutions For Class 11 Maths Chapter 8 Exercise 8.2 are written in a stepwise manner for all questions. Other chapters of Class 11 Mathematics are:

In Class 11 Mathematics there are a total of 16 chapters. They are:

1. Chapter 1 Sets
2. Chapter 2 Relations and Functions
3. Chapter 3 Trigonometric Functions
4. Chapter 4 Principles of Mathematical Induction
5. Chapter 5 Complex numbers and Quadratic equations
6. Chapter 6 Linear Inequalities
7. Chapter 7 Permutation and Combinations
8. Chapter 8 Binomial Theorem
9. Chapter 9 Sequences and Series
10. Chapter 10 Straight lines
11. Chapter 11 Conic Sections
12. Chapter 12 Introduction to Three Dimensional Geometry
13. Chapter 13 Limits and Derivatives
14. Chapter 14 Mathematical Reasoning
15. Chapter 15 Statistics
16. Chapter 16 Probability

All the above-mentioned chapters are important to score good marks in the Class 11 Mathematics examination. Students should solve the NCERT exercises of all the above chapters. They can take the help of NCERT Solutions in case of any difficulty. Learners should also make notes of the chapters for revision before the exam.

Q.1 Find the coefficient of x5 in (x + 3)8.

Ans.

$\begin{array}{l}{\text{The (r + 1)}}^{\text{th}}{\text{term of the expansion (x + y)}}^{\text{n}}\text{is given by}\\ {\text{T}}_{\text{r\hspace{0.17em}+1}}={\text{}}^{\text{n}}{\text{C}}_{\text{r}}{{\text{x}}^{\text{n}}}^{-\text{r}}{\text{y}}^{\text{r}}\text{.}\\ \text{In​ the expansion of}{\left(\text{x}+\text{3}\right)}^{\text{8}},\\ {\text{T}}_{\text{r\hspace{0.17em}+1}}={\text{}}^{\text{8}}{\text{C}}_{\text{r}}{{\text{x}}^{\text{8}}}^{-\text{r}}{\text{\hspace{0.17em}3}}^{\text{r}}\\ \mathrm{For}{\text{the coefficient of x}}^{\text{5}},\text{}\\ \text{let \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}8}-\mathrm{r}=5\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{r}=8-5=3\\ \mathrm{So},{\text{the coefficient of x}}^{\text{5}}\\ ={\text{\hspace{0.17em}}}^{\text{8}}{\text{C}}_{\text{3}}{\text{\hspace{0.17em}3}}^{\text{3}}\\ =27×\frac{8!}{3!5!}\\ =27×\frac{8×7×6×5!}{3×2×1×5!}\\ =1512\end{array}$

Q.2 Find the coefficient of a5b7 in (a – 2b)12.

Ans.

$\begin{array}{l}{\text{The (r + 1)}}^{\text{th}}{\text{term of the expansion (x + y)}}^{\text{n}}\text{is given by}\\ {\text{T}}_{\text{r\hspace{0.17em}+1}}={\text{}}^{\text{n}}{\text{C}}_{\text{r}}{{\text{x}}^{\text{n}}}^{-\text{r}}{\text{y}}^{\text{r}}\text{.}\\ \text{In​ the expansion of}{\left(\text{a}-\text{2b}\right)}^{\text{12}},\\ {\text{T}}_{\text{r\hspace{0.17em}+1}}={\text{}}^{\text{12}}{\text{C}}_{\text{r}}{{\text{a}}^{\text{12}}}^{-\text{r}}\text{\hspace{0.17em}}{\left(-\text{2b}\right)}^{\text{r}}\\ \mathrm{For}{\text{the coefficient of a}}^{\text{5}}{\text{b}}^{\text{7}},\text{}\\ \text{let 12}-\mathrm{r}=5\\ ⇒\mathrm{r}=12-5\\ \begin{array}{l}=7\\ \mathrm{So},{\text{the coefficient of a}}^{\text{5}}{\text{b}}^{\text{7}}\\ ={\text{\hspace{0.17em}}}^{\text{12}}{\text{C}}_{\text{7}}\text{\hspace{0.17em}}{\left(-2\right)}^{\text{7}}\\ =-128×\frac{12!}{7!5!}\\ =-128×\frac{12×11×10×9×8×7!}{7!\text{\hspace{0.17em}}×5×4×3×2×1}\\ =-101376\end{array}\end{array}$

Q.3 Write the general term in the expansion of (x2 – y)6.

Ans.

$\begin{array}{l}\mathrm{The}\text{general term of}{\left({\text{x}}^{\text{2}}–\text{y}\right)}^{\text{6}}\text{\hspace{0.17em}}\mathrm{is}\\ {\mathrm{T}}_{\mathrm{r}+1}={\text{\hspace{0.17em}}}^{6}{\mathrm{C}}_{\mathrm{r}}{\left({\mathrm{x}}^{2}\right)}^{6-\mathrm{r}}{\left(-\mathrm{y}\right)}^{\mathrm{r}}\\ =\text{\hspace{0.17em}}{\left(-1\right)}^{\mathrm{r}}{\text{\hspace{0.17em}}}^{6}{\mathrm{C}}_{\mathrm{r}}{\left(\mathrm{x}\right)}^{12-2\mathrm{r}}.{\mathrm{y}}^{\mathrm{r}}\end{array}$

Q.4 Find the 4th term in the expansion of (x – 2y)12.

Ans.

$\begin{array}{l}\mathrm{Let}{\text{​ 4}}^{\text{th}}\text{term in the expression of}{\left(\text{x}-\text{2y}\right)}^{\text{12}}{\text{be T}}_{\text{4}}\text{,\hspace{0.17em}then}\\ {\text{T}}_{\text{4}}={\mathrm{T}}_{3+1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\text{\hspace{0.17em}}}^{12}{\mathrm{C}}_{3}{\mathrm{x}}^{12-3}{\left(-2\mathrm{y}\right)}^{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\text{\hspace{0.17em}}}^{12}{\mathrm{C}}_{3}{\mathrm{x}}^{9}{\left(-2\right)}^{3}{\mathrm{y}}^{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\text{\hspace{0.17em}}8×\frac{12!}{3!\text{\hspace{0.17em}\hspace{0.17em}}9!}{\mathrm{x}}^{9}{\mathrm{y}}^{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\text{\hspace{0.17em}}8×\frac{12×11×10×9!}{3×2×1×\text{\hspace{0.17em}\hspace{0.17em}}9!}{\mathrm{x}}^{9}{\mathrm{y}}^{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\text{\hspace{0.17em}}1760\text{\hspace{0.17em}}{\mathrm{x}}^{9}{\mathrm{y}}^{3}\end{array}$

Q.5

${\mathbf{\text{Find the 13}}}^{\mathbf{\text{th}}}\mathbf{\text{term in the expansion of\hspace{0.17em}\hspace{0.17em}}}{\left(\text{9x}-\frac{\text{1}}{\text{3}\sqrt{\text{x}}}\right)}^{\mathbf{\text{18}}}\mathbf{\text{,\hspace{0.17em}x}}\mathbf{\ne }\mathbf{\text{0.}}$

Ans.

$\begin{array}{l}{\text{Let​ 13}}^{\text{th}}\text{term in the expression of}{\left(9\mathrm{x}-\frac{1}{3\sqrt{\mathrm{x}}}\right)}^{18}{\text{be T}}_{\text{13}}\text{,\hspace{0.17em}then}\\ {\text{T}}_{\text{13}}={\mathrm{T}}_{12+1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\text{\hspace{0.17em}}}^{18}{\mathrm{C}}_{12}{\left(9\mathrm{x}\right)}^{18-12}{\left(-\frac{1}{3\sqrt{\mathrm{x}}}\right)}^{12}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}={\text{\hspace{0.17em}}}^{18}{\mathrm{C}}_{12}\left({9}^{6}{\mathrm{x}}^{6}\right){\left(-\frac{1}{3}\right)}^{12}\left(\frac{1}{{\mathrm{x}}^{6}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\text{\hspace{0.17em}}\frac{{9}^{6}}{{3}^{12}}×\frac{18!}{12!\text{\hspace{0.17em}\hspace{0.17em}}6!}{\mathrm{x}}^{6}\left(\frac{1}{{\mathrm{x}}^{6}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\text{\hspace{0.17em}}\frac{18×17×16×15×14×13}{6×5×4×3×2}×\frac{{\mathrm{x}}^{6}}{{\mathrm{x}}^{6}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=18564\end{array}$

Q.6

$\mathbf{Write}{\mathbf{\text{the general term in the expansion of (x}}}^{\mathbf{2}}\mathbf{-}\mathbf{yx}{\mathbf{\text{)}}}^{\mathbf{12}}\mathbf{,}\mathbf{\text{}}\mathbf{x}\mathbf{\ne }\mathbf{0}\mathbf{.}$

Ans.

$\begin{array}{l}\mathrm{The}\text{general term of}{\left({\text{x}}^{\text{2}}-\text{yx}\right)}^{\text{12}}\text{\hspace{0.17em}}\mathrm{is}\\ {\mathrm{T}}_{\mathrm{r}+1}={\text{\hspace{0.17em}}}^{12}{\mathrm{C}}_{\mathrm{r}}{\left({\mathrm{x}}^{2}\right)}^{12-\mathrm{r}}{\left(-\mathrm{yx}\right)}^{\mathrm{r}}\\ =\text{\hspace{0.17em}}{\left(-1\right)}^{\mathrm{r}}{\text{\hspace{0.17em}}}^{12}{\mathrm{C}}_{\mathrm{r}}{\left(\mathrm{x}\right)}^{24-2\mathrm{r}}.{\mathrm{x}}^{\mathrm{r}}.{\mathrm{y}}^{\mathrm{r}}\\ =\text{\hspace{0.17em}}{\left(-1\right)}^{\mathrm{r}}{\text{\hspace{0.17em}}}^{12}{\mathrm{C}}_{\mathrm{r}}{\left(\mathrm{x}\right)}^{24-\mathrm{r}}.{\mathrm{y}}^{\mathrm{r}}\end{array}$

Q.7

$\mathbf{\text{Find the middle term in the expansion of\hspace{0.17em}\hspace{0.17em}}}{\left(\text{3}-\frac{{\text{x}}^{\text{3}}}{\text{6}}\right)}^{\mathbf{\text{7}}}\mathbf{\text{.}}$

Ans.

$\begin{array}{l}{\left(3-\frac{{\mathrm{x}}^{3}}{6}\right)}^{7}\\ \mathrm{Here},\text{\hspace{0.17em}}\mathrm{n}=7\left(\mathrm{odd}\right)\\ \mathrm{So},\text{middle terms in expansion}\\ ={\left(\frac{7+1}{2}\right)}^{\mathrm{th}}\text{\hspace{0.17em}}\mathrm{term}\text{and}{\left(\frac{7+1}{2}+1\right)}^{\mathrm{th}}\text{\hspace{0.17em}}\mathrm{term}\\ ={4}^{\mathrm{th}}{\text{​ term and 5}}^{\text{th}}\text{term}\\ {\text{Then,\hspace{0.17em}\hspace{0.17em}T}}_{\text{4}}={\mathrm{T}}_{3+1}\\ ={\text{\hspace{0.17em}}}^{7}{\mathrm{C}}_{3}.{3}^{7-3}.{\left(-\frac{{\mathrm{x}}^{3}}{6}\right)}^{3}\\ ={\text{\hspace{0.17em}}}^{7}{\mathrm{C}}_{3}.{3}^{4}.{\left(-\frac{1}{6}\right)}^{3}{\mathrm{x}}^{9}\\ =\text{\hspace{0.17em}}-\text{\hspace{0.17em}}\frac{35×27×3}{36×6}{\mathrm{x}}^{9}\\ =-\frac{105}{8}{\mathrm{x}}^{9}\\ \mathrm{and}{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}T}}_{\text{5}}={\mathrm{T}}_{4+1}\end{array}$

$\begin{array}{l}={\text{\hspace{0.17em}}}^{7}{\mathrm{C}}_{4}.{3}^{7-4}.{\left(-\frac{{\mathrm{x}}^{3}}{6}\right)}^{4}\\ ={\text{\hspace{0.17em}}}^{7}{\mathrm{C}}_{4}.{3}^{3}{\left(-\frac{1}{6}\right)}^{4}{\mathrm{x}}^{12}\\ =\text{\hspace{0.17em}\hspace{0.17em}}\frac{35×27}{36×6×6}{\mathrm{x}}^{12}\\ =\frac{35}{48}{\mathrm{x}}^{12}\\ \mathrm{Thus},\text{the middle terms of the expansion are}-\frac{105}{8}{\text{x}}^{\text{9}}\text{\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}\frac{35}{48}{\mathrm{x}}^{12}.\end{array}$

Q.8

$\mathbf{\text{Find the middle term in the expansion of\hspace{0.17em}\hspace{0.17em}}}{\left(\frac{\text{x}}{\text{3}}\text{+9y}\right)}^{\mathbf{\text{10}}}\mathbf{\text{.}}$

Ans.

$\begin{array}{l}\mathrm{Given}:\text{ }{\left(\frac{\text{x}}{3}+9\text{y}\right)}^{10}\\ \mathrm{Here},\text{ n}=10\text{ }\left(\mathrm{even}\right)\\ \mathrm{The}\mathrm{middle}\mathrm{term}\mathrm{in}\mathrm{the}\text{expansion}={\left(\frac{10}{2}+1\right)}^{\mathrm{th}}\text{term}\\ \text{ }={\text{6}}^{\text{th}}\text{ term}\\ {\text{Then, T}}_{\text{6}}={\text{T}}_{5+1}\\ ={\text{ }}^{10}{\text{C}}_{5}{\left(\frac{\text{x}}{3}\right)}^{10-5}{\left(9\text{y}\right)}^{5}\\ \text{ }=\frac{10!}{5!\left(10-5\right)!}×\frac{{\text{x}}^{5}}{{3}^{5}}×{9}^{5}{\text{y}}^{5}\\ \text{ }=\frac{10×9×8×7×6×5!}{5×4×3×2×1×5!}×\frac{{\text{x}}^{5}}{{3}^{5}}×{3}^{10}{\text{y}}^{5}\\ \text{ }=6×42×9×27{\text{x}}^{5}{\text{y}}^{4}\\ \text{ }=61236{\text{ x}}^{5}{\text{y}}^{5}\\ \mathrm{Thus},\text{the middle term of given expression is}61236{\text{ x}}^{5}{\text{y}}^{5}.\end{array}$

Q.9 In the expansion of (1 + a)m+n, prove that coefficients of am and an are equal.

Ans.

$\begin{array}{l}\mathrm{The}\text{given expression is}{\left(\text{1}+\text{a}\right)}^{\text{m}+\text{n}}.\\ \mathrm{The}\text{general term}\left({\mathrm{T}}_{\mathrm{r}+1}\right)={\text{\hspace{0.17em}}}^{\mathrm{m}+\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}{\mathrm{a}}^{\mathrm{r}}\text{\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Putting}\text{r}=\mathrm{m}\text{in equation}\left(\mathrm{i}\right),\text{\hspace{0.17em}}\mathrm{we}\text{\hspace{0.17em}}\mathrm{get}\\ \mathrm{Coefficient}{\text{of a}}^{\text{m}}={\text{\hspace{0.17em}}}^{\mathrm{m}+\mathrm{n}}{\mathrm{C}}_{\mathrm{m}}\\ \mathrm{Putting}\text{r}=\mathrm{n}\text{in equation}\left(\mathrm{i}\right),\text{\hspace{0.17em}}\mathrm{we}\text{\hspace{0.17em}}\mathrm{get}\\ \mathrm{Coefficient}{\text{of a}}^{\text{n}}={\text{\hspace{0.17em}}}^{\mathrm{m}+\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}\\ ={\text{\hspace{0.17em}}}^{\mathrm{m}+\mathrm{n}}{\mathrm{C}}_{\mathrm{m}+\mathrm{n}-\text{\hspace{0.17em}}\mathrm{n}}\left[\because {\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}={\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-\mathrm{r}}\right]\\ ={\text{\hspace{0.17em}}}^{\mathrm{m}+\mathrm{n}}{\mathrm{C}}_{\mathrm{m}}\\ \mathrm{Thus},\text{}\\ {\text{the coefficient of a}}^{\text{m}}={\text{the coefficient of a}}^{\text{n}}.\end{array}$

Q.10 The coefficients of the (r – 1)th, rth and (r + 1)th terms in the expansion of (x + 1)n are in the ratio 1 : 3 : 5. Find n and r.

Ans.

$\begin{array}{l}\text{The coefficients of the}{\left(\text{r}-\text{1}\right)}^{\text{th}},{\text{r}}^{\text{th}}\text{and}\mathrm{}\text{}{\left(\text{r}+\text{1}\right)}^{\text{th}}\text{terms in the}\\ \text{expansion of}{\left(\text{x}+\text{1}\right)}^{\text{n}}\mathrm{}\text{are in the ratio 1}:\text{3}:\text{5.Then,}\\ {\text{\hspace{0.17em}T}}_{\text{r}-\text{1}}={\text{T}}_{\left(\text{r}-\text{2}\right)\text{\hspace{0.17em}+1}}\\ ={\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}-2}{\mathrm{x}}^{\mathrm{n}-\left(\mathrm{r}-2\right)}\\ =\frac{\mathrm{n}!}{\left(\mathrm{r}-2\right)!\left(\mathrm{n}-\mathrm{r}+2\right)!}{\mathrm{x}}^{\left(\mathrm{n}-\mathrm{r}+2\right)}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}T}}_{\text{r}}={\text{T}}_{\left(\text{r}-\text{1}\right)\text{\hspace{0.17em}+1}}\\ ={\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{\left(\mathrm{r}-1\right)}{\mathrm{x}}^{\mathrm{n}-\left(\mathrm{r}-1\right)}\\ =\frac{\mathrm{n}!}{\left(\mathrm{r}-1\right)!\left(\mathrm{n}-\mathrm{r}+1\right)!}{\mathrm{x}}^{\left(\mathrm{n}-\mathrm{r}+1\right)}\\ {\text{\hspace{0.17em}\hspace{0.17em}T}}_{\text{r}+\text{1}}={\text{T}}_{\mathrm{r}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{\hspace{0.17em}}={\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}{\mathrm{x}}^{\mathrm{n}-\mathrm{r}}\\ =\frac{\mathrm{n}!}{\mathrm{r}!\left(\mathrm{n}-\mathrm{r}\right)!}{\mathrm{x}}^{\left(\mathrm{n}-\mathrm{r}\right)}\\ \mathrm{Since},\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}}\frac{\mathrm{Coefficient}\text{of}{\mathrm{T}}_{\mathrm{r}-1}}{\mathrm{Coefficient}\text{of}{\mathrm{T}}_{\mathrm{r}}}=\frac{\frac{\mathrm{n}!}{\left(\mathrm{r}-2\right)!\left(\mathrm{n}-\mathrm{r}+2\right)!}}{\frac{\mathrm{n}!}{\left(\mathrm{r}-1\right)!\left(\mathrm{n}-\mathrm{r}+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}}\frac{1}{3}=\frac{\left(\mathrm{r}-1\right)!\left(\mathrm{n}-\mathrm{r}+1\right)!}{\left(\mathrm{r}-2\right)!\left(\mathrm{n}-\mathrm{r}+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{1}{3}=\frac{\left(\mathrm{r}-1\right).\left(\mathrm{r}-2\right)!\left(\mathrm{n}-\mathrm{r}+1\right)!}{\left(\mathrm{r}-2\right)!\left(\mathrm{n}-\mathrm{r}+2\right).\left(\mathrm{n}-\mathrm{r}+1\right)!}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\end{array}$

$\begin{array}{l}\text{\hspace{0.17em}}\frac{1}{3}=\frac{\left(\mathrm{r}-1\right)}{\left(\mathrm{n}-\mathrm{r}+2\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{n}-\mathrm{r}+2=3\text{\hspace{0.17em}}\mathrm{r}-3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{n}-4\text{\hspace{0.17em}}\mathrm{r}=-\text{\hspace{0.17em}}5...\left(\mathrm{i}\right)\\ \text{\hspace{0.17em}}\frac{\mathrm{Coefficient}\text{of}{\mathrm{T}}_{\mathrm{r}}}{\mathrm{Coefficient}\text{of}{\mathrm{T}}_{\mathrm{r}+1}}=\frac{\frac{\mathrm{n}!}{\left(\mathrm{r}-1\right)!\left(\mathrm{n}-\mathrm{r}+1\right)!}}{\frac{\mathrm{n}!}{\mathrm{r}!\left(\mathrm{n}-\mathrm{r}\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{3}{5}=\frac{\mathrm{r}.\left(\mathrm{r}-1\right)!\left(\mathrm{n}-\mathrm{r}\right)!}{\left(\mathrm{r}-1\right)!\left(\mathrm{n}-\mathrm{r}+1\right).\left(\mathrm{n}-\mathrm{r}\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{3}{5}=\frac{\mathrm{r}\text{\hspace{0.17em}}}{\left(\mathrm{n}-\mathrm{r}+1\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3\text{\hspace{0.17em}}\mathrm{n}-3\text{\hspace{0.17em}}\mathrm{r}+3=5\mathrm{r}\\ \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}}3\text{\hspace{0.17em}}\mathrm{n}-8\text{\hspace{0.17em}}\mathrm{r}=-3...\left(\mathrm{ii}\right)\\ \mathrm{On}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{solving}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we get}\\ \text{n}=\text{7 and r}=\text{3.}\end{array}$

Q.11 Prove that the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n – 1.

Ans.

$\begin{array}{l}\mathrm{Given}\text{expressions are:\hspace{0.17em}\hspace{0.17em}}{\left(\text{1}+\text{x}\right)}^{\text{2n}}\text{and}{\left(\text{1}+\text{x}\right)}^{\text{2n-1}}\\ \mathrm{For}\text{\hspace{0.17em}\hspace{0.17em}}{\left(\text{1}+\text{x}\right)}^{\text{2n}}:\\ \mathrm{Since},{\text{T}}_{\text{r+1}}={\text{\hspace{0.17em}}}^{2\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}{\mathrm{x}}^{\mathrm{r}}\\ \mathrm{Putting}\text{r}=\text{n, we get}\\ {\text{coefficient of x}}^{\text{n}}={\text{\hspace{0.17em}}}^{2\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\end{array}$

$\begin{array}{l}=\frac{2\mathrm{n}!}{\mathrm{n}!\left(2\mathrm{n}-\mathrm{n}\right)!}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2\mathrm{n}!}{\mathrm{n}!\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{n}!}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2\mathrm{n}\left(2\mathrm{n}-1\right)!}{\mathrm{n}\left(\mathrm{n}-1\right)!\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{n}!}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2×\frac{\left(2\mathrm{n}-1\right)!}{\left(\mathrm{n}-1\right)!\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{n}!}\\ \mathrm{For}\text{\hspace{0.17em}\hspace{0.17em}}{\left(\text{1}+\text{x}\right)}^{\text{2n-1}}:\\ \mathrm{Since},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{T}}_{\mathrm{r}+1}={\text{\hspace{0.17em}}}^{2\mathrm{n}-1}{\mathrm{C}}_{\mathrm{r}}{\mathrm{x}}^{\mathrm{r}}\\ \mathrm{Putting}\text{r}=\text{n, we get}\\ {\text{coefficient of x}}^{\text{n}}={\text{\hspace{0.17em}}}^{2\mathrm{n}-1}{\mathrm{C}}_{\mathrm{n}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(2\mathrm{n}-1\right)!}{\mathrm{n}!\left(2\mathrm{n}-1-\mathrm{n}\right)!}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(2\mathrm{n}-1\right)!}{\mathrm{n}!\left(\mathrm{n}-1\right)!}\\ \mathrm{Thus},{\text{coefficient of x}}^{\text{n}}\text{in}{\left(1+\mathrm{x}\right)}^{\mathrm{n}}\text{is double of the coefficient}\\ {\text{of x}}^{\text{n}}\text{in}{\left(1+\mathrm{x}\right)}^{2\mathrm{n}-1}.\end{array}$

Q.12 Find a positive value of m for which the coefficient of x2 in the expansion (1 + x)m is 6.

Ans.

$\begin{array}{l}{\text{Given: Coefficient of x}}^{\text{2}}\text{in the expansion of}{\left(\text{1}+\text{x}\right)}^{\text{m}}\text{\hspace{0.17em}}\mathrm{is}\text{}6.\\ {\text{The (r + 1)}}^{\text{th}}\text{term of the expansion}{\left(\text{1}+\text{x}\right)}^{\text{m}}\text{is given by}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}T}}_{\text{r+1}}={\text{\hspace{0.17em}}}^{\text{m}}{\text{C}}_{\text{r}}{\text{x}}^{\text{r}}\\ \text{Putting r}=\text{2, we get}\\ {\text{coefficient of x}}^{\text{2}}={\text{\hspace{0.17em}}}^{\text{m}}{\text{C}}_{\text{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}}6=\frac{\mathrm{m}!}{2!\left(\mathrm{m}-2\right)!}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}6=\frac{\mathrm{m}\left(\mathrm{m}-1\right)\left(\mathrm{m}-2\right)!}{2\left(\mathrm{m}-2\right)!}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}12=\mathrm{m}\left(\mathrm{m}-1\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4×3=\mathrm{m}\left(\mathrm{m}-1\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4×\left(4-1\right)=\mathrm{m}\left(\mathrm{m}-1\right)\\ ⇒\mathrm{m}=4\end{array}$

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### 1. Where can students and parents find the NCERT Solutions For Class 11 Maths Chapter 8 Exercise 8.2?

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Students can use the NCERT Solutions For Class 11 Maths Chapter 8 Exercise 8.2 for making short notes, and for revising important formulas a few days before the exams. They can also use the NCERT Solutions For Class 11 Maths Chapter 8 Exercise 8.2 for self-assessment of their preparation. After this self-assessment, they can identify their weak areas and work on them.

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