# NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem (Ex 8.1) Exercise 8.1

Students give their Class 10 board examinations and step up a grade to Class 11. Out of all the academic years, Class 11 is one of the most crucial for them. Students from the Central Board of Secondary Education (CBSE) affiliated schools select their core stream of subjects in Class 11. Students select out of the three streams that are Science stream, Commerce stream and Humanities. Some schools allocate the streams to the students on the basis of the marks that they have obtained in their Class 10 board examinations, whereas some schools allow the students to choose the stream they want to study irrespective of the marks that they have obtained in their Class 10 board examinations.

Students who select the Commerce stream study subjects such as Business Studies, Accounts, Economics, etc. Students who select the Humanities stream study subjects such as Political Science, History, Sociology, etc. Students who opt for the Science stream study the subjects such as Physics, Chemistry, Biology, and Mathematics. They are selecting the subjects that they want to study, and this choice can be very difficult for them. It is a very crucial decision to make, as the subject stream that they select will most likely determine their future career. In this crucial period, parents should help and support their children. With the support of one’s parents, students can make the right decision for their future. A few aspects that students should keep in mind while selecting their subject stream are their interest in the subjects and what they want to become in the future.

Before preparing for the Mathematics exam, students should know their syllabus. A good performance in exams depends on studying notes and NCERT solutions. In preparation for board exams, students should be familiar with the accepted method of answering questions. There are probably many different methods and formulas that can be used to solve a single question in Mathematics. However, if the method does not meet the criteria for their marking system, the school or board may not consider it acceptable. In this case, students may lose marks even if they provide the correct answer. The NCERT Solutions For Class 11 Maths Chapter 8 Exercise 8.1, will help students who are struggling with this problem.

After Class 10, the difficulty level increases in Classes 11 and 12. These classes are the most important academic years of the students’ life. As the difficulty level increases, students have to study more and put in more effort into their respective subjects. Students especially have to put more effort as with each year CBSE papers are becoming more difficult. They should take their Class 11 studies seriously as they lay the foundation for Class 12. If they do not study properly for Class 11, they will find it difficult to understand what is being taught in Class 12.

For those students who have opted for Science subjects such as Physics, Chemistry And Mathematics, Mathematics becomes a very important subject. Studying mathematics is not only important for scoring good marks but also because it helps in studying the concepts and theories behind Physics and Chemistry too. Class 11 Mathematics is very important to understand the basic concepts of Class 12 studies. Students should prepare for the subject from an exam point of view. If prepared fully, mathematics is a very scoring subject and helps students obtain an overall good percentage. It is a very practical subject. Its applications can be found all around the world. Studying mathematics also develops a student’s reasoning and analytical skills. Mathematics can be used in day-to-day life. Learners can find the NCERT Solutions For Class 11 Maths Chapter 8 Exercise 8.1 on Extramarks’ website and mobile application to study for Chapter 8 Class 11 Maths Exercise 8.1.

## NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem (Ex 8.1) Exercise 8.1

Mathematics Class 11 Chapter 8 is named – Binomial Theorem. This chapter is very important for the students to score well in the exam. Students can find the  NCERT Solutions For Class 11 Maths Chapter 8 Exercise 8.1 on the Extramarks mobile application and website. The  NCERT Solutions For Class 11 Maths Chapter 8 Exercise 8.1 can be downloaded in PDF format. Students can use the  NCERT Solutions For Class 11 Maths Chapter 8 Exercise 8.1 to find answers to difficult questions. They can also refer to the  NCERT Solutions For Class 11 Maths Chapter 8 Exercise 8.1 to revise important theorems and formulas.

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

Students and parents can download the NCERT Solutions For Class 11 Maths Chapter 8 Exercise 8.1 from the Extramarks’ website and mobile application. Parents can also use the NCERT Solutions For Class 11 Maths Chapter 8 Exercise 8.1 to help their children in their studies.

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### NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Exercise 8.1

For the Class 11 examinations, all subjects and all chapters are important. To avoid making mistakes in the Class 11 Mathematics paper, it is recommended that students practice all the exercise questions in the back of the chapters. If students face any problems in solving the exercises, they can refer to the NCERT Solutions For Class 11 Maths Chapter 8 Exercise 8.1, from the Extramarks website and mobile application. The NCERT Solutions For Class 11 Maths Chapter 8 Exercise 8.1, will provide them with accurate solutions for all the questions from Exercise 8.1 Class 11. Learners can download the NCERT Solutions For Class 11 Maths Chapter 8 Exercise 8.1 in PDF format for their convenience.

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Q.1 Expand the expression: (1 – 2x)5.

Ans.

$\begin{array}{l}\mathrm{Since},\text{\hspace{0.17em}}{\left(1-\mathrm{x}\right)}^{\mathrm{n}}={\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{0}-{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{1}\mathrm{x}+{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{2}{\mathrm{x}}^{2}-...+\text{\hspace{0.17em}}{\left(-1\right)}^{\mathrm{n}\text{\hspace{0.17em}}}{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(\text{1}-\text{2x}\right)}^{\text{5}}={\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{0}-{\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{1}\left(2\mathrm{x}\right)+{\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{2}{\left(2\mathrm{x}\right)}^{2}-{\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{3}{\left(2\mathrm{x}\right)}^{3}+{\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{4}{\left(2\mathrm{x}\right)}^{4}\text{\hspace{0.17em}}-{\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{5}{\left(2\mathrm{x}\right)}^{5}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1-10\mathrm{x}+40{\mathrm{x}}^{2}-80{\mathrm{x}}^{3}+80{\mathrm{x}}^{4}-32{\mathrm{x}}^{5}\end{array}$

Q.2

$\mathbf{\text{Expand the expression:}}{\left(\frac{\text{2}}{\text{x}}\text{–}\frac{\text{x}}{\text{2}}\right)}^{\mathbf{\text{5}}}\mathbf{\text{.}}$

Ans.

$\begin{array}{l}\text{Since,\hspace{0.17em}}{\left(\mathrm{a}-\mathrm{b}\right)}^{\mathrm{n}}={\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{0}{\mathrm{a}}^{\mathrm{n}}-{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{1}{\mathrm{a}}^{\mathrm{n}-1}\mathrm{b}+{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{2}{\mathrm{a}}^{\mathrm{n}-2}{\mathrm{b}}^{2}-...+\text{\hspace{0.17em}}{\left(-1\right)}^{\mathrm{n}\text{\hspace{0.17em}}}{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}\text{\hspace{0.17em}}{\mathrm{b}}^{\mathrm{n}}\\ \text{So,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(\frac{2}{\mathrm{x}}-\frac{\mathrm{x}}{2}\right)}^{5}={\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{0}{\left(\frac{2}{\mathrm{x}}\right)}^{5}-{\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{1}{\left(\frac{2}{\mathrm{x}}\right)}^{4}\left(\frac{\mathrm{x}}{2}\right)+{\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{2}{\left(\frac{2}{\mathrm{x}}\right)}^{3}{\left(\frac{\mathrm{x}}{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}\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}}}^{5}{\mathrm{C}}_{3}{\left(\frac{2}{\mathrm{x}}\right)}^{2}{\left(\frac{\mathrm{x}}{2}\right)}^{3}+{\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{4}\left(\frac{2}{\mathrm{x}}\right){\left(\frac{\mathrm{x}}{2}\right)}^{4}\text{\hspace{0.17em}}-{\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{5}{\left(\frac{\mathrm{x}}{2}\right)}^{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}}=\text{\hspace{0.17em}}\left(1\right)\left(\frac{32}{{\mathrm{x}}^{5}}\right)-\text{\hspace{0.17em}}\left(5\right)\left(\frac{16}{{\mathrm{x}}^{4}}\right)\left(\frac{\mathrm{x}}{2}\right)+\text{\hspace{0.17em}}\left(10\right)\left(\frac{8}{{\mathrm{x}}^{3}}\right)\left(\frac{{\mathrm{x}}^{2}}{4}\right)\\ \text{\hspace{0.17em}}-\text{\hspace{0.17em}}\left(10\right)\left(\frac{4}{{\mathrm{x}}^{2}}\right)\left(\frac{{\mathrm{x}}^{3}}{8}\right)+\text{\hspace{0.17em}}\left(5\right)\left(\frac{2}{\mathrm{x}}\right)\left(\frac{{\mathrm{x}}^{4}}{16}\right)\text{\hspace{0.17em}}-\text{\hspace{0.17em}}\left(1\right)\left(\frac{{\mathrm{x}}^{5}}{32}\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{32}{{\mathrm{x}}^{5}}-\frac{40}{{\mathrm{x}}^{3}}+\frac{20}{\mathrm{x}}-5\mathrm{x}+\frac{5}{8}{\mathrm{x}}^{3}-\frac{{\mathrm{x}}^{5}}{32}\end{array}$

Q.3 Expand the expression: (2x – 3)6.

Ans.

$\begin{array}{l}\mathrm{Since},\text{\hspace{0.17em}}{\left(\mathrm{a}-\mathrm{b}\right)}^{\mathrm{n}}={\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{0}{\mathrm{a}}^{\mathrm{n}}-{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{1}{\mathrm{a}}^{\mathrm{n}-1}\mathrm{b}+{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{2}{\mathrm{a}}^{\mathrm{n}-2}{\mathrm{b}}^{2}-...+\text{\hspace{0.17em}}{\left(-1\right)}^{\mathrm{n}\text{\hspace{0.17em}}}{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}\text{\hspace{0.17em}}{\mathrm{b}}^{\mathrm{n}}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(\text{2x}-\text{3}\right)}^{\text{6}}={\text{\hspace{0.17em}\hspace{0.17em}}}^{6}{\mathrm{C}}_{0}{\left(2\mathrm{x}\right)}^{6}-{\text{\hspace{0.17em}}}^{6}{\mathrm{C}}_{1}{\left(2\mathrm{x}\right)}^{5}\left(3\right)+{\text{\hspace{0.17em}}}^{6}{\mathrm{C}}_{2}{\left(2\mathrm{x}\right)}^{4}{\left(3\right)}^{2}-{\text{\hspace{0.17em}}}^{6}{\mathrm{C}}_{3}{\left(2\mathrm{x}\right)}^{3}{\left(3\right)}^{3}\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}+{\text{\hspace{0.17em}}}^{6}{\mathrm{C}}_{4}{\left(2\mathrm{x}\right)}^{2}{\left(3\right)}^{4}-{\text{\hspace{0.17em}}}^{6}{\mathrm{C}}_{5}{\left(2\mathrm{x}\right)}^{1}{\left(3\right)}^{5}+{\text{\hspace{0.17em}}}^{6}{\mathrm{C}}_{6}{\left(2\mathrm{x}\right)}^{0}{\left(3\right)}^{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}\hspace{0.17em}}=\left(1\right){2}^{6}{\mathrm{x}}^{6}-6×{2}^{5}{\mathrm{x}}^{5}×3+15×{2}^{4}{\mathrm{x}}^{4}×{3}^{2}-20×{2}^{3}{\mathrm{x}}^{3}×{3}^{3}\\ +15×{2}^{2}{\mathrm{x}}^{2}×{3}^{4}-6×{2}^{1}{\mathrm{x}}^{1}×{3}^{5}+\left(1\right){3}^{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}\hspace{0.17em}}=64\text{\hspace{0.17em}}{\mathrm{x}}^{6}-576{\mathrm{x}}^{5}+2160\text{\hspace{0.17em}}{\mathrm{x}}^{4}-4320{\mathrm{x}}^{3}+4860{\mathrm{x}}^{2}\\ -2916\mathrm{x}+729\end{array}$

Q.4 Expand the expression:

${\left(\frac{\text{x}}{\text{3}}\text{+}\frac{\text{1}}{\text{x}}\right)}^{\mathbf{\text{5}}}\mathbf{\text{.}}$

Ans.

$\begin{array}{l}\mathrm{Since},\text{\hspace{0.17em}}{\left(\mathrm{a}+\mathrm{b}\right)}^{\mathrm{n}}={\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{0}{\mathrm{a}}^{\mathrm{n}}+{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{1}{\mathrm{a}}^{\mathrm{n}-1}\mathrm{b}+{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{2}{\mathrm{a}}^{\mathrm{n}-2}{\mathrm{b}}^{2}+...+{\text{\hspace{0.17em}\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}\text{\hspace{0.17em}}{\mathrm{b}}^{\mathrm{n}}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(\frac{\mathrm{x}}{3}+\frac{1}{\mathrm{x}}\right)}^{5}={\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{0}{\left(\frac{\mathrm{x}}{3}\right)}^{5}+{\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{1}{\left(\frac{\mathrm{x}}{3}\right)}^{4}\left(\frac{1}{\mathrm{x}}\right)+{\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{2}{\left(\frac{\mathrm{x}}{3}\right)}^{3}{\left(\frac{1}{\mathrm{x}}\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}\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}}}^{5}{\mathrm{C}}_{3}{\left(\frac{\mathrm{x}}{3}\right)}^{2}{\left(\frac{1}{\mathrm{x}}\right)}^{3}+{\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{4}\left(\frac{\mathrm{x}}{3}\right){\left(\frac{1}{\mathrm{x}}\right)}^{4}\text{\hspace{0.17em}}+{\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{5}{\left(\frac{1}{\mathrm{x}}\right)}^{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}}=\text{\hspace{0.17em}}\left(1\right)\left(\frac{{\mathrm{x}}^{5}}{243}\right)+\text{\hspace{0.17em}}\left(5\right)\left(\frac{{\mathrm{x}}^{4}}{81}\right)\left(\frac{1}{\mathrm{x}}\right)+\text{\hspace{0.17em}}\left(10\right)\left(\frac{{\mathrm{x}}^{3}}{27}\right)\left(\frac{1}{{\mathrm{x}}^{2}}\right)\\ \text{\hspace{0.17em}}+\text{\hspace{0.17em}}\left(10\right)\left(\frac{{\mathrm{x}}^{2}}{9}\right)\left(\frac{1}{{\mathrm{x}}^{3}}\right)+\text{\hspace{0.17em}}\left(5\right)\left(\frac{\mathrm{x}}{3}\right)\left(\frac{1}{{\mathrm{x}}^{4}}\right)\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\left(1\right)\left(\frac{1}{{\mathrm{x}}^{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}}=\frac{{\mathrm{x}}^{5}}{243}+\frac{5{\mathrm{x}}^{3}}{81}+\frac{10\mathrm{x}}{27}+\frac{10}{9\mathrm{x}}+\frac{5}{3{\mathrm{x}}^{3}}+\frac{1}{{\mathrm{x}}^{5}}\end{array}$

Q.5

$\mathbf{\text{Expand the expression:}}{\left(\text{x+}\frac{\text{1}}{\text{x}}\right)}^{\mathbf{\text{6}}}\mathbf{\text{.}}$

Ans.

$\begin{array}{l}\mathrm{Since},\text{\hspace{0.17em}}{\left(\mathrm{a}+\mathrm{b}\right)}^{\mathrm{n}}={\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{0}{\mathrm{a}}^{\mathrm{n}}+{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{1}{\mathrm{a}}^{\mathrm{n}-1}\mathrm{b}+{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{2}{\mathrm{a}}^{\mathrm{n}-2}{\mathrm{b}}^{2}+...+{\text{\hspace{0.17em}}}^{\mathrm{n}\text{\hspace{0.17em}}}{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}\text{\hspace{0.17em}}{\mathrm{b}}^{\mathrm{n}}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(\text{x}+\frac{1}{\mathrm{x}}\right)}^{\text{6}}={\text{\hspace{0.17em}\hspace{0.17em}}}^{6}{\mathrm{C}}_{0}{\left(\mathrm{x}\right)}^{6}+{\text{\hspace{0.17em}}}^{6}{\mathrm{C}}_{1}{\left(\mathrm{x}\right)}^{5}\left(\frac{1}{\mathrm{x}}\right)+{\text{\hspace{0.17em}}}^{6}{\mathrm{C}}_{2}{\left(\mathrm{x}\right)}^{4}{\left(\frac{1}{\mathrm{x}}\right)}^{2}+{\text{\hspace{0.17em}}}^{6}{\mathrm{C}}_{3}{\left(\mathrm{x}\right)}^{3}{\left(\frac{1}{\mathrm{x}}\right)}^{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+{\text{\hspace{0.17em}}}^{6}{\mathrm{C}}_{4}{\left(\mathrm{x}\right)}^{2}{\left(\frac{1}{\mathrm{x}}\right)}^{4}+{\text{\hspace{0.17em}}}^{6}{\mathrm{C}}_{5}{\left(\mathrm{x}\right)}^{1}{\left(\frac{1}{\mathrm{x}}\right)}^{5}+{\text{\hspace{0.17em}}}^{6}{\mathrm{C}}_{6}{\left(\mathrm{x}\right)}^{0}{\left(\frac{1}{\mathrm{x}}\right)}^{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}\hspace{0.17em}}=\left(1\right){\mathrm{x}}^{6}+6×{\mathrm{x}}^{4}+15×{\mathrm{x}}^{2}+20\text{\hspace{0.17em}}+15×\frac{1}{{\mathrm{x}}^{2}}+6×\frac{1}{{\mathrm{x}}^{4}}+\left(1\right)\frac{1}{{\mathrm{x}}^{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}\hspace{0.17em}}={\mathrm{x}}^{6}+6{\mathrm{x}}^{4}+15\text{\hspace{0.17em}}{\mathrm{x}}^{2}+20+\frac{15}{{\mathrm{x}}^{2}}+\frac{6}{{\mathrm{x}}^{4}}+\frac{1}{{\mathrm{x}}^{6}}\end{array}$

Q.6 Using binomial theorem, evaluate the following: (96)3

Ans.

$\begin{array}{l}\mathrm{Since},\text{\hspace{0.17em}}\\ {\left(\mathrm{a}-\mathrm{b}\right)}^{\mathrm{n}}={\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{0}{\mathrm{a}}^{\mathrm{n}}-{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{1}{\mathrm{a}}^{\mathrm{n}-1}\mathrm{b}+{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{2}{\mathrm{a}}^{\mathrm{n}-2}{\mathrm{b}}^{2}-...+\text{\hspace{0.17em}}{\left(-1\right)}^{\mathrm{n}\text{\hspace{0.17em}}}{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}\text{\hspace{0.17em}}{\mathrm{b}}^{\mathrm{n}}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}}{\left(96\right)}^{3}={\left(100-4\right)}^{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\text{\hspace{0.17em}}}^{3}{\mathrm{C}}_{0}{\left(100\right)}^{3}-{\text{\hspace{0.17em}}}^{3}{\mathrm{C}}_{1}{\left(100\right)}^{3-1}\left(4\right)+{\text{\hspace{0.17em}}}^{3}{\mathrm{C}}_{2}{\left(100\right)}^{3-2}{\left(4\right)}^{2}\\ -{\text{\hspace{0.17em}}}^{3}{\mathrm{C}}_{3}{\left(100\right)}^{3-3}{\left(4\right)}^{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1000000-3×10000×4+3×100×16-1×1×64\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1000000-120000+4800-64\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=884736\end{array}$

Q.7 Using binomial theorem, evaluate the following: (102)5.

Ans.

$\begin{array}{l}\mathrm{Since},\text{\hspace{0.17em}}{\left(\mathrm{a}+\mathrm{b}\right)}^{\mathrm{n}}={\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{0}{\mathrm{a}}^{\mathrm{n}}+{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{1}{\mathrm{a}}^{\mathrm{n}-1}\mathrm{b}+{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{2}{\mathrm{a}}^{\mathrm{n}-2}{\mathrm{b}}^{2}+...+{\text{\hspace{0.17em}\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}\text{\hspace{0.17em}}{\mathrm{b}}^{\mathrm{n}}\\ \because \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(102\right)}^{5}={\left(100+2\right)}^{5}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}}{\left(100+2\right)}^{5}={\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{0}{\left(100\right)}^{5}+{\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{1}{\left(100\right)}^{4}\left(2\right)+{\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{2}{\left(100\right)}^{3}{\left(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}\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}}}^{5}{\mathrm{C}}_{3}{\left(100\right)}^{2}{\left(2\right)}^{3}+{\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{4}\left(100\right){\left(2\right)}^{4}\text{\hspace{0.17em}}+{\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{5}{\left(2\right)}^{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}}=\text{\hspace{0.17em}}\left(1\right)\left(10000000000\right)+\text{\hspace{0.17em}}\left(5\right)\left(100000000\right)\left(2\right)\\ +\text{\hspace{0.17em}}\left(10\right)\left(1000000\right)\left(4\right)+\text{\hspace{0.17em}}\left(10\right)\left(10000\right)\left(8\right)\\ +\text{\hspace{0.17em}}\left(5\right)\left(100\right)\left(16\right)\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\left(1\right)\left(32\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}}=10000000000+1000000000+40000000+800000\\ \text{\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}}8000+32\\ \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}}=11040808032\end{array}$

Q.8 Using binomial theorem, evaluate the following: (101)4

Ans.

$\begin{array}{l}\mathrm{Since},\text{\hspace{0.17em}}{\left(\mathrm{a}+\mathrm{b}\right)}^{\mathrm{n}}={\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{0}{\mathrm{a}}^{\mathrm{n}}+{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{1}{\mathrm{a}}^{\mathrm{n}-1}\mathrm{b}+{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{2}{\mathrm{a}}^{\mathrm{n}-2}{\mathrm{b}}^{2}+...+{\text{\hspace{0.17em}\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}\text{\hspace{0.17em}}{\mathrm{b}}^{\mathrm{n}}\\ \because \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(101\right)}^{4}={\left(100+2\right)}^{4}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}}{\left(100+1\right)}^{4}={\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{0}{\left(100\right)}^{4}+{\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{1}{\left(100\right)}^{3}\left(1\right)+{\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{2}{\left(100\right)}^{2}{\left(1\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}\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}}}^{4}{\mathrm{C}}_{3}{\left(100\right)}^{1}{\left(1\right)}^{3}+{\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{4}{\left(1\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}}=\text{\hspace{0.17em}}\left(1\right)\left(100000000\right)+\text{\hspace{0.17em}}\left(4\right)\left(1000000\right)\left(1\right)\\ +\text{\hspace{0.17em}}\left(6\right)\left(10000\right)\left(1\right)+\text{\hspace{0.17em}}\left(4\right)\left(100\right)\left(1\right)+\text{\hspace{0.17em}}\left(1\right)\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}}=100000000+4000000+60000+400+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}}=104060401\end{array}$

Q.9 Using Binomial Theorem, indicate which number is larger (1.1)10000 or 1000.

Ans.

$\begin{array}{l}{\left(\text{1.1}\right)}^{\text{10000}}\text{=}{\left(\text{1+0.1}\right)}^{\text{10000}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=1+10000}\left(\text{0.1}\right)\text{+\hspace{0.17em}higher\hspace{0.17em}powers\hspace{0.17em}}...\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=1+1000+\hspace{0.17em}\hspace{0.17em}higher\hspace{0.17em}powers\hspace{0.17em}}...\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}>1001\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\text{Leaving higher power terms}\right)\\ \because \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}1001>1000}\\ \text{So,\hspace{0.17em}}{\left(\text{1.1}\right)}^{\text{10000}}\text{>1000.}\end{array}$

Q.10

${\mathbf{\text{Find (a + b)}}}^{\mathbf{\text{4}}}\mathbf{-}{\mathbf{\text{(a}}\mathbf{-}\mathbf{\text{b)}}}^{\mathbf{\text{4}}}\mathbf{\text{. Hence, evaluate}}{\left(\sqrt{\text{3}}\text{+}\sqrt{\text{2}}\right)}^{\mathbf{\text{4}}}\mathbf{-}{\left(\sqrt{\text{3}}-\sqrt{\text{2}}\right)}^{\mathbf{\text{4}}}\mathbf{\text{.}}$

Ans.

$\begin{array}{l}\mathrm{Since},\text{\hspace{0.17em}}{\left(\mathrm{a}+\mathrm{b}\right)}^{\mathrm{n}}={\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{0}{\mathrm{a}}^{\mathrm{n}}+{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{1}{\mathrm{a}}^{\mathrm{n}-1}\mathrm{b}+{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{2}{\mathrm{a}}^{\mathrm{n}-2}{\mathrm{b}}^{2}+...+{\text{\hspace{0.17em}\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}\text{\hspace{0.17em}}{\mathrm{b}}^{\mathrm{n}}\\ {\text{(a + b)}}^{\text{4}}-{\text{(a – b)}}^{\text{4}}=\text{\hspace{0.17em}}{\left(}^{4}{\mathrm{C}}_{0}{\mathrm{a}}^{4}+{\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{1}{\mathrm{a}}^{3}\mathrm{b}+{\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{2}{\mathrm{a}}^{2}{\mathrm{b}}^{2}+{\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{3}{\mathrm{a}}^{1}{\mathrm{b}}^{3}+{\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{4}{\mathrm{b}}^{4}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-{\left(}^{4}{\mathrm{C}}_{0}{\mathrm{a}}^{4}-{\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{1}{\mathrm{a}}^{3}\mathrm{b}+{\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{2}{\mathrm{a}}^{2}{\mathrm{b}}^{2}-{\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{3}{\mathrm{a}}^{1}{\mathrm{b}}^{3}+{\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{4}{\mathrm{b}}^{4}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{0}{\mathrm{a}}^{4}+{\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{1}{\mathrm{a}}^{3}\mathrm{b}+{\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{2}{\mathrm{a}}^{2}{\mathrm{b}}^{2}+{\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{3}{\mathrm{a}}^{1}{\mathrm{b}}^{3}+{\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{4}{\mathrm{b}}^{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{-}^{4}{\mathrm{C}}_{0}{\mathrm{a}}^{4}+{\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{1}{\mathrm{a}}^{3}\mathrm{b}-{\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{2}{\mathrm{a}}^{2}{\mathrm{b}}^{2}+{\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{3}{\mathrm{a}}^{1}{\mathrm{b}}^{3}-{\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{4}{\mathrm{b}}^{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{\hspace{0.17em}\hspace{0.17em}}2\left({\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{1}{\mathrm{a}}^{3}\mathrm{b}+{\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{3}{\mathrm{a}}^{1}{\mathrm{b}}^{3}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{\hspace{0.17em}\hspace{0.17em}}2\left(4{\mathrm{a}}^{3}\mathrm{b}+4{\mathrm{ab}}^{3}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8\left({\mathrm{a}}^{3}\mathrm{b}+{\mathrm{ab}}^{3}\right)\\ \mathrm{Putting}\text{}\mathrm{a}=\sqrt{3}\text{​}\mathrm{and}\text{}\mathrm{b}=\sqrt{2},\text{}\mathrm{we}\text{}\mathrm{get}\\ {\left(\sqrt{3}+\sqrt{2}\right)}^{4}-{\left(\sqrt{3}-\sqrt{2}\right)}^{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8\left\{{\left(\sqrt{3}\right)}^{3}\left(\sqrt{2}\right)+\left(\sqrt{3}\right){\left(\sqrt{2}\right)}^{3}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8\left\{3\sqrt{6}+2\sqrt{6}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8\left(5\sqrt{6}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=40\sqrt{6}\end{array}$

Q.11

$\begin{array}{l}{\mathbf{\text{Find (x + 1)}}}^{\mathbf{\text{6}}}\mathbf{\text{+(x}}\mathbf{-}{\mathbf{\text{1)}}}^{\mathbf{\text{6}}}\mathbf{\text{. Hence or otherwise evaluate}}\\ \mathbf{\text{}}{\left(\sqrt{\text{2}}\text{+1}\right)}^{\mathbf{\text{6}}}\mathbf{-}{\left(\sqrt{\text{2}}-\text{1}\right)}^{\mathbf{\text{6}}}\mathbf{\text{.}}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Since},\text{\hspace{0.17em}}{\left(\mathrm{a}+\mathrm{b}\right)}^{\mathrm{n}}={\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{0}{\mathrm{a}}^{\mathrm{n}}+{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{1}{\mathrm{a}}^{\mathrm{n}-1}\mathrm{b}+{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{2}{\mathrm{a}}^{\mathrm{n}-2}{\mathrm{b}}^{2}+...+{\text{\hspace{0.17em}}}^{\mathrm{n}\text{\hspace{0.17em}}}{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}\text{\hspace{0.17em}}{\mathrm{b}}^{\mathrm{n}}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}{\left(\mathrm{x}+1\right)}^{\text{6}}={\text{\hspace{0.17em}\hspace{0.17em}}}^{6}{\mathrm{C}}_{0}{\left(\mathrm{x}\right)}^{6}+{\text{\hspace{0.17em}}}^{6}{\mathrm{C}}_{1}{\left(\mathrm{x}\right)}^{5}+{\text{\hspace{0.17em}}}^{6}{\mathrm{C}}_{2}{\left(\mathrm{x}\right)}^{4}+{\text{\hspace{0.17em}}}^{6}{\mathrm{C}}_{3}{\left(\mathrm{x}\right)}^{3}\text{\hspace{0.17em}}+{\text{\hspace{0.17em}}}^{6}{\mathrm{C}}_{4}{\left(\mathrm{x}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+{\text{\hspace{0.17em}}}^{6}{\mathrm{C}}_{5}{\left(\mathrm{x}\right)}^{1}+{\text{\hspace{0.17em}}}^{6}{\mathrm{C}}_{6}{\left(\mathrm{x}\right)}^{0}\\ \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}}=\left(1\right){\mathrm{x}}^{6}+6{\mathrm{x}}^{5}+15\text{\hspace{0.17em}}{\mathrm{x}}^{4}+20{\mathrm{x}}^{3}\text{\hspace{0.17em}}+15\text{\hspace{0.17em}}{\mathrm{x}}^{2}+6\mathrm{x}+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}}{\left(\mathrm{x}-1\right)}^{\text{6}}={\text{\hspace{0.17em}\hspace{0.17em}}}^{6}{\mathrm{C}}_{0}{\left(\mathrm{x}\right)}^{6}-{\text{\hspace{0.17em}}}^{6}{\mathrm{C}}_{1}{\left(\mathrm{x}\right)}^{5}+{\text{\hspace{0.17em}}}^{6}{\mathrm{C}}_{2}{\left(\mathrm{x}\right)}^{4}-{\text{\hspace{0.17em}}}^{6}{\mathrm{C}}_{3}{\left(\mathrm{x}\right)}^{3}\text{\hspace{0.17em}}+{\text{\hspace{0.17em}}}^{6}{\mathrm{C}}_{4}{\left(\mathrm{x}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-{\text{\hspace{0.17em}}}^{6}{\mathrm{C}}_{5}{\left(\mathrm{x}\right)}^{1}+{\text{\hspace{0.17em}}}^{6}{\mathrm{C}}_{6}{\left(\mathrm{x}\right)}^{0}\\ \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}}=\left(1\right){\mathrm{x}}^{6}-6{\mathrm{x}}^{5}+15\text{\hspace{0.17em}}{\mathrm{x}}^{4}-20{\mathrm{x}}^{3}\text{\hspace{0.17em}}+15\text{\hspace{0.17em}}{\mathrm{x}}^{2}-6\mathrm{x}+1\\ {\text{(x + 1)}}^{\text{6}}{\text{+(x – 1)}}^{\text{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}\hspace{0.17em}}=2{\mathrm{x}}^{6}+30\text{\hspace{0.17em}}{\mathrm{x}}^{4}+30\text{\hspace{0.17em}}{\mathrm{x}}^{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}}=2\left({\mathrm{x}}^{6}+15\text{\hspace{0.17em}}{\mathrm{x}}^{4}+15\text{\hspace{0.17em}}{\mathrm{x}}^{2}+1\right)\\ \mathrm{Putting}\text{x}=\sqrt{2},\text{we get}\\ {\left(\sqrt{2}+1\right)}^{6}-{\left(\sqrt{2}-1\right)}^{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}\hspace{0.17em}}=2\left\{{\left(\sqrt{2}\right)}^{6}+15\text{\hspace{0.17em}}{\left(\sqrt{2}\right)}^{4}+15\text{\hspace{0.17em}}{\left(\sqrt{2}\right)}^{2}+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}}=2\left(8+60+30+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}}=2\left(99\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}}=198\end{array}$

Q.12 Show that 9n+1 – 8n – 9 is divisible by 64, whenever n is a positive integer.

Ans.

$\begin{array}{l}{9}^{\text{n}+1}={\left(1+8\right)}^{\text{n}+1}\\ \text{ }=1+\left(\text{n}+1\right)8+\frac{\left(\text{n}+1\right)\text{n}}{2!}{8}^{2}+\frac{\left(\text{n}+1\right)\text{n}\left(\text{n}-1\right)}{3!}{8}^{3}+...\\ ⇒\text{ }{9}^{\text{n}+1}-1-8\text{n}-8=\frac{\left(\text{n}+1\right)\text{n}}{2!}{8}^{2}+\frac{\left(\text{n}+1\right)\text{n}\left(\text{n}-1\right)}{3!}{8}^{3}+...\\ \text{ }{9}^{\text{n}+1}-8\text{n}-9={8}^{2}\left\{\frac{\left(\text{n}+1\right)\text{n}}{2!}+\frac{\left(\text{n}+1\right)\text{n}\left(\text{n}-1\right)}{3!}8+...\right\}\\ \text{ }{9}^{\text{n}+1}-8\text{n}-9=64\left\{\frac{\left(\text{n}+1\right)\text{n}}{2!}+\frac{\left(\text{n}+1\right)\text{n}\left(\text{n}-1\right)}{3!}8+...\right\}\\ {9}^{\text{n}+1}-8\text{n}-9=\mathrm{multiple}\text{of 64, because n is an integer.}\\ ⇒\text{ }{9}^{\text{n}+1}-8\text{n}-9\text{is divisible by 64.}\end{array}$

Q.13

$\mathbf{\text{Prove that}}\sum _{\text{r=0}}^{\text{n}}{\text{3}}^{\text{r}}{\text{\hspace{0.17em}}}^{\text{n}}{\text{C}}_{\text{r}}{\text{\hspace{0.17em}=4}}^{\text{n}}\text{.}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\sum _{\mathrm{r}=0}^{\mathrm{n}}{3}^{\mathrm{r}}{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}\text{\hspace{0.17em}}={3}^{0}{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{0}+{3}^{1}{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{1}+{3}^{2}{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{2}+{3}^{3}{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{3}+{3}^{4}{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{4}+...\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{0}\text{\hspace{0.17em}}.{3}^{0}{+}^{\mathrm{n}}{\mathrm{C}}_{1}\text{\hspace{0.17em}}.{3}^{1}{+}^{\mathrm{n}}{\mathrm{C}}_{2}\text{\hspace{0.17em}}.{3}^{2}{+}^{\mathrm{n}}{\mathrm{C}}_{3}\text{\hspace{0.17em}}.{3}^{3}+...\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left(1+3\right)}^{\mathrm{n}}\left[\begin{array}{l}\because {\left(1+\mathrm{x}\right)}^{\mathrm{n}}={\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{0}\text{\hspace{0.17em}}.{\mathrm{x}}^{0}+{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{1}\text{\hspace{0.17em}}.{\mathrm{x}}^{1}+{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{2}\text{\hspace{0.17em}}.{\mathrm{x}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+{\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{3}\text{\hspace{0.17em}}.{\mathrm{x}}^{3}+...\end{array}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={4}^{\mathrm{n}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

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