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# NCERT Solutions Class 11 Maths Chapter 7 Binomial Theorem

Last updated date: 20th Sep 2024
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## NCERT Solutions for Maths Binomial Theorem Class 11 Chapter 7 - FREE PDF Download

By referring to NCERT Solutions for Binomial Theorem Class 11 Maths Chapter 7, students are able to Understand the topics covered in this chapter in detail according to the CBSE Class 11 Maths Syllabus. The binomial theorem is defined as the process of algebraically expanding the power of sums of two or more binomials. Coefficients of binomial terms in the process of expansion are referred to as binomial coefficients.

Table of Content
1. NCERT Solutions for Maths Binomial Theorem Class 11 Chapter 7 - FREE PDF Download
2. Glance on Maths Chapter 7 Class 11 - Binomial Theorem
3. Access Exercise wise NCERT Solutions for Chapter 7 Maths Class 11
4. Exercises Under NCERT Solutions for Class 11 Maths Chapter 7 Binomial Theorem
5. Access NCERT Solutions for Class 11 Maths Chapter 7 – Binomial Theorem
5.1Exercise 7.1
5.2Miscellaneous Exercise:
6. Overview of Deleted Syllabus for CBSE Class 11 Maths Binomial Theorem
7. Class 11 Maths Chapter 7: Exercises Breakdown
8. Other Study Material for CBSE Class 11 Maths Chapter 7
9. Chapter-Specific NCERT Solutions for Class 11 Maths
FAQs

Our subject experts at Vedantu have solved all the sums and explained all the topics covered in this chapter according to the CBSE guidelines in these NCERT Solutions Class 11 Maths. You can download these solutions from Vedantu for free.

## Glance on Maths Chapter 7 Class 11 - Binomial Theorem

• Chapter 7 Maths Class 11 explains the expansion of binomial expressions using the binomial theorem.

• A binomial expression is an algebraic expression containing two terms.

• The binomial theorem provides a formula for expanding expressions raised to any positive integer power.

• Binomial coefficients are the numerical factors in terms of a binomial expansion.

• Pascal's Triangle is a triangular array of numbers that provides the binomial coefficients.

• The general term of a binomial expansion represents a specific term in the expanded form.

• The middle term is the central term(s) in the binomial expansion when the exponent is even or odd.

• There are two exercises including Miscellaneous Exercise (20 fully solved questions) in Class 11  Maths Chapter 7  Binomial Theorem

## Access Exercise wise NCERT Solutions for Chapter 7 Maths Class 11

 Current Syllabus Exercises of Class 11 Maths Chapter 7 NCERT Solutions of Class 10 Maths Binomial Theorem Exercise 7.1 NCERT Solutions of Class 10 Maths Binomial Theorem Miscellaneous Exercise
Competitive Exams after 12th Science

## Exercises Under NCERT Solutions for Class 11 Maths Chapter 7 Binomial Theorem

### Exercise 7.1: This exercise consists of 14 questions and introduces students to the concept of binomial theorem. The questions are based on various concepts such as expanding binomial expressions, finding the middle terms of a binomial expansion, and solving problems related to the binomial theorem. Students will learn how to expand binomial expressions using the binomial theorem.

Miscellaneous Exercise: This exercise consists of 6 questions and covers a variety of topics related to binomial theorem. The questions are based on various concepts such as solving problems related to the number of terms in a binomial expansion, finding the sum of the terms in a binomial expansion, and solving word problems related to binomial theorem. This exercise will help students to revise and reinforce the concepts learned in the previous exercises.

Overall, the exercises in NCERT Solutions for Class 11 Maths Chapter  7 – Binomial Theorem are designed to help students understand and apply the concepts of binomial theorem in various scenarios. The solutions to these exercises are provided in the textbook, which will help students to check their answers and understand the concepts better.

## Access NCERT Solutions for Class 11 Maths Chapter 7 – Binomial Theorem

### Exercise 7.1

1. Expand the expression ${\left( {1 - 2x} \right)^5}$.

Ans. By using Binomial Theorem, the expression ${\left( {1 - 2x} \right)^5}$ can be expanded as

$\begin{gathered} {\left( {1 - 2x} \right)^5} = {}^5{C_0}{\left( 1 \right)^5} - {}^5{C_1}{\left( 1 \right)^4}\left( {2x} \right) + {}^5{C_2}{\left( 1 \right)^3}{\left( {2x} \right)^2} - {}^5{C_3}{\left( 1 \right)^2}{\left( {2x} \right)^3} + {}^5{C_4}{\left( 1 \right)^1}{\left( {2x} \right)^4} \\ - {}^5{C_5}{\left( {2x} \right)^5} \\ = 1 - 5\left( {2x} \right) + 10\left( {4{x^2}} \right) - 10\left( {8{x^3}} \right) + 5\left( {16{x^4}} \right) - 32{x^5} \\ = 1 - 10x + 40{x^2} - 80{x^3} + 80{x^4} - 32{x^5} \\ \end{gathered}$

2. Expand the expression ${\left( {\frac{2}{x} - \frac{x}{2}} \right)^5}$.

Ans. By using Binomial Theorem, the expression ${\left( {\frac{2}{x} - \frac{x}{2}} \right)^5}$ can be expanded as

$\begin{gathered} {\left( {\frac{2}{x} - \frac{x}{2}} \right)^5} = {}^5{C_0}{\left( {\frac{2}{x}} \right)^5} - {}^5{C_1}{\left( {\frac{2}{x}} \right)^4}\left( {\frac{x}{2}} \right) + {}^5{C_2}{\left( {\frac{2}{x}} \right)^3}{\left( {\frac{x}{2}} \right)^2} - {}^5{C_3}{\left( {\frac{2}{x}} \right)^2}{\left( {\frac{x}{2}} \right)^3} + {}^5{C_4}{\left( {\frac{2}{x}} \right)^1}{\left( {\frac{x}{2}} \right)^4} \\ - {}^5{C_5}{\left( {\frac{x}{2}} \right)^5} \\ = \frac{{32}}{{{x^5}}} - 5\left( {\frac{{16}}{{{x^4}}}} \right)\left( {\frac{x}{2}} \right) + 10\left( {\frac{8}{{{x^3}}}} \right)\left( {\frac{{{x^2}}}{4}} \right) - 10\left( {\frac{4}{{{x^2}}}} \right)\left( {\frac{{{x^3}}}{8}} \right) + 5\left( {\frac{2}{x}} \right)\left( {\frac{{{x^4}}}{{16}}} \right) - \frac{{{x^5}}}{{32}} \\ = \frac{{32}}{{{x^5}}} - \frac{{40}}{{{x^3}}} + \frac{{20}}{x} - 5x + \frac{5}{8}{x^3} - \frac{{{x^5}}}{{32}} \\ \end{gathered}$

3. Expand the expression ${\left( {2x - 3} \right)^6}$.

Ans. By using Binomial Theorem, the expression ${\left( {2x - 3} \right)^6}$ can be expanded as

$\begin{gathered} {\left( {2x - 3} \right)^6} = {}^6{C_0}{\left( {2x} \right)^6} - {}^6{C_1}{\left( {2x} \right)^5}\left( 3 \right) + {}^6{C_2}{\left( {2x} \right)^4}{\left( 3 \right)^2} - {}^6{C_3}{\left( {2x} \right)^3}{\left( 3 \right)^3} + {}^6{C_4}{\left( {2x} \right)^2}{\left( 3 \right)^4} \\ - {}^6{C_5}\left( {2x} \right){\left( 3 \right)^5} + {}^6{C_6}{\left( 3 \right)^6} \\ = 64{x^6} - 6\left( {32{x^5}} \right)\left( 3 \right) + 15\left( {16{x^4}} \right)\left( 9 \right) - 20\left( {8{x^3}} \right)\left( {27} \right) + 15\left( {4{x^2}} \right)\left( {81} \right) \\ - 6\left( {2x} \right)\left( {243} \right) + 729 \\ = 64{x^6} - 576{x^5} + 2160{x^4} - 4320{x^3} + 4860{x^2} - 2916x + 729 \\ \end{gathered}$

4. Expand the expression ${\left( {\frac{x}{3} + \frac{1}{x}} \right)^5}$.

Ans. By using Binomial Theorem, the expression ${\left( {\frac{x}{3} + \frac{1}{x}} \right)^5}$ can be expanded as

$\begin{gathered} {\left( {\frac{x}{3} + \frac{1}{x}} \right)^5} = {}^5{C_0}{\left( {\frac{x}{3}} \right)^5} + {}^5{C_1}{\left( {\frac{x}{3}} \right)^4}\left( {\frac{1}{x}} \right) + {}^5{C_2}{\left( {\frac{x}{3}} \right)^3}{\left( {\frac{1}{x}} \right)^2} + {}^5{C_3}{\left( {\frac{x}{3}} \right)^2}{\left( {\frac{1}{x}} \right)^3} + {}^5{C_4}{\left( {\frac{x}{3}} \right)^1}{\left( {\frac{1}{x}} \right)^4} \\ + {}^5{C_5}{\left( {\frac{1}{x}} \right)^5} \\ = \frac{{{x^5}}}{{243}} + 5\left( {\frac{{{x^4}}}{{81}}} \right)\left( {\frac{1}{x}} \right) + 10\left( {\frac{{{x^3}}}{{27}}} \right)\left( {\frac{1}{{{x^2}}}} \right) + 10\left( {\frac{{{x^2}}}{9}} \right)\left( {\frac{1}{{{x^3}}}} \right) + 5\left( {\frac{x}{3}} \right)\left( {\frac{1}{{{x^4}}}} \right) + \frac{1}{{{x^5}}} \\ = \frac{{{x^5}}}{{243}} + \frac{{5{x^3}}}{{81}} + \frac{{10x}}{{27}} + \frac{{10}}{{9x}} + \frac{5}{{3{x^3}}} + \frac{1}{{{x^5}}} \\ \end{gathered}$

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

Ans. By using Binomial Theorem, the expression ${\left( {x + \frac{1}{x}} \right)^6}$ can be expanded as

$\begin{gathered} {\left( {x + \frac{1}{x}} \right)^6} = {}^6{C_0}{\left( x \right)^6} + {}^6{C_1}{\left( x \right)^5}\left( {\frac{1}{x}} \right) + {}^6{C_2}{\left( x \right)^4}{\left( {\frac{1}{x}} \right)^2} + {}^6{C_3}{\left( x \right)^3}{\left( {\frac{1}{x}} \right)^3} + {}^6{C_4}{\left( x \right)^2}{\left( {\frac{1}{x}} \right)^4} \\ + {}^6{C_5}\left( x \right){\left( {\frac{1}{x}} \right)^5} + {}^6{C_6}{\left( {\frac{1}{x}} \right)^6} \\ = {x^6} + 6\left( {{x^5}} \right)\left( {\frac{1}{x}} \right) + 15\left( {{x^4}} \right)\left( {\frac{1}{{{x^2}}}} \right) + 20\left( {{x^3}} \right)\left( {\frac{1}{{{x^3}}}} \right) + 15\left( {{x^2}} \right)\left( {\frac{1}{{{x^4}}}} \right) + 6\left( x \right)\left( {\frac{1}{{{x^5}}}} \right) + \frac{1}{{{x^6}}} \\ = {x^6} + 6{x^4} + 15{x^2} + 20 + \frac{{15}}{{{x^2}}} + \frac{6}{{{x^4}}} + \frac{1}{{{x^6}}} \\ \end{gathered}$

6. Using Binomial Theorem, evaluate ${\left( {96} \right)^3}$.

Ans. 96 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, the binomial theorem can be applied.

It can be written that, $96 = 100 - 4$

$\begin{gathered} {\left( {96} \right)^3} = {\left( {100 - 4} \right)^3} \\ = {}^3{C_0}{\left( {100} \right)^3} - {}^3{C_1}{\left( {100} \right)^2}\left( 4 \right) + {}^3{C_2}\left( {100} \right){\left( 4 \right)^2} - {}^3{C_3}{\left( 4 \right)^3} \\ = 1000000 - 3\left( {10000} \right)\left( 4 \right) + 3\left( {100} \right)\left( {16} \right) - 64 \\ = 1000000 - 120000 + 4800 - 64 \\ = 884736 \\ \end{gathered}$

7. Using Binomial Theorem, evaluate ${\left( {102} \right)^5}$.

Ans. 102 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, the binomial theorem can be applied.

It can be written that, $102 = 100 + 2$

$\begin{gathered} {\left( {102} \right)^5} = {\left( {100 + 2} \right)^5} \\ = {}^5{C_0}{\left( {100} \right)^5} + {}^5{C_1}{\left( {100} \right)^4}\left( 2 \right) + {}^5{C_2}{\left( {100} \right)^3}{\left( 2 \right)^2} + {}^5{C_3}{\left( {100} \right)^2}{\left( 2 \right)^3} + {}^5{C_4}\left( {100} \right){\left( 2 \right)^4} \\ + {}^5{C_5}{\left( 2 \right)^5} \\ = 10000000000 + 5\left( {100000000} \right)\left( 2 \right) + 10\left( {1000000} \right)\left( 4 \right) + 10\left( {10000} \right)\left( 8 \right) \\ + 5\left( {100} \right)\left( {16} \right) + 32 \\ = 10000000000 + 1000000000 + 40000000 + 80000 + 8000 + 32 \\ = 11040808032 \\ \end{gathered}$

8. Using Binomial Theorem, evaluate ${\left( {101} \right)^4}$.

Ans. 101 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, the binomial theorem can be applied.

It can be written that, $101 = 100 + 1$

$\begin{gathered} {\left( {101} \right)^4} = {\left( {100 + 1} \right)^4} \\ = {}^4{C_0}{\left( {100} \right)^4} + {}^4{C_1}{\left( {100} \right)^3}\left( 1 \right) + {}^4{C_2}{\left( {100} \right)^2}{\left( 1 \right)^2} + {}^4{C_3}\left( {100} \right){\left( 1 \right)^3} + {}^4{C_4}{\left( 1 \right)^4} \\ = 100000000 + 4\left( {1000000} \right) + 6\left( {10000} \right) + 4\left( {100} \right) + 1 \\ = 100000000 + 4000000 + 60000 + 400 + 1 \\ = 104060401 \\ \end{gathered}$

9. Using Binomial Theorem, evaluate ${\left( {99} \right)^5}$.

Ans. 99 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, the binomial theorem can be applied.

It can be written that, $99 = 100 - 1$

$\begin{gathered} {\left( {99} \right)^5} = {\left( {100 - 1} \right)^5} \\ = {}^5{C_0}{\left( {100} \right)^5} - {}^5{C_1}{\left( {100} \right)^4}\left( 1 \right) + {}^5{C_2}{\left( {100} \right)^3}{\left( 1 \right)^2} - {}^5{C_3}{\left( {100} \right)^2}{\left( 1 \right)^3} + {}^5{C_4}\left( {100} \right){\left( 1 \right)^4} \\ - {}^5{C_5}{\left( 1 \right)^5} \\ = 10000000000 - 5\left( {100000000} \right) - 10\left( {1000000} \right) - 10\left( {10000} \right) + 5\left( {100} \right) - 1 \\ = 10000000000 - 500000000 - 10000000 - 100000 + 500 - 1 \\ = 9509900499 \\ \end{gathered}$

10. Using Binomial Theorem, indicate which number is larger ${\left( {1.1} \right)^{10000}}$ or $1000$.

Ans. By splitting 1.1 and then applying the Binomial Theorem, the first few terms of ${\left( {1.1} \right)^{10000}}$ be obtained as

${\left( {1.1} \right)^{10000}}$ be obtained as $\begin{gathered} {\left( {1.1} \right)^{10000}} = {\left( {1 + 0.1} \right)^{10000}} \\ = {}^{10000}{C_0} + {}^{10000}{C_1}\left( {1.1} \right) + {\text{Other positive terms}} \\ = 1 + 10000 \times 1.1 + {\text{Other positive terms}} \\ = 1 + 11000 + {\text{Other positive terms}} \\ > 1000 \\ \end{gathered} Hence, \[{\left( {1.1} \right)^{10000}} > 1000$

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

Ans. Using Binomial Theorem, the expressions, ${\left( {a + b} \right)^4}$ and ${\left( {a - b} \right)^4}$ , can be expanded as

$\begin{gathered} {\left( {a + b} \right)^4} = {}^4{C_0}{a^4} + {}^4{C_1}{a^3}b + {}^4{C_2}{a^2}{b^2} + {}^4{C_3}a{b^3} + {}^4{C_4}{b^4} \\ {\left( {a - b} \right)^4} = {}^4{C_0}{a^4} - {}^4{C_1}{a^3}b + {}^4{C_2}{a^2}{b^2} - {}^4{C_3}a{b^3} + {}^4{C_4}{b^4} \\ \end{gathered}$

Therefore,

$\begin{gathered} {\left( {a + b} \right)^4} - {\left( {a - b} \right)^4} = {}^4{C_0}{a^4} + {}^4{C_1}{a^3}b + {}^4{C_2}{a^2}{b^2} + {}^4{C_3}a{b^3} + {}^4{C_4}{b^4} - \\ \left[ {{}^4{C_0}{a^4} - {}^4{C_1}{a^3}b + {}^4{C_2}{a^2}{b^2} - {}^4{C_3}a{b^3} + {}^4{C_4}{b^4}} \right] \\ = 2\left( {{}^4{C_1}{a^3}b + {}^4{C_3}a{b^3}} \right) \\ = 2\left( {4{a^3}b + 4a{b^3}} \right) \\ = 8ab\left( {{a^2} + {b^2}} \right) \\ \end{gathered}$

By putting $a = \sqrt 3$ and $b = \sqrt 2$, we obtain

$\begin{gathered} {\left( {\sqrt 3 + \sqrt 2 } \right)^4} - {\left( {\sqrt 3 - \sqrt 2 } \right)^4} = 8\left( {\sqrt 3 } \right)\left( {\sqrt 2 } \right)\left[ {{{\left( {\sqrt 3 } \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}} \right] \\ = 8\sqrt 6 \left( {3 + 2} \right) \\ = 40\sqrt 6 \\ \end{gathered}$

12. Find ${\left( {x + 1} \right)^6} + {\left( {x - 1} \right)^6}$. Hence or otherwise evaluate ${\left( {\sqrt 2 + 1} \right)^6} + {\left( {\sqrt 2 - 1} \right)^6}$.

Ans. Using Binomial Theorem, the expressions, ${\left( {x + 1} \right)^6}$ and ${\left( {x - 1} \right)^6}$ , can be expanded as

$\begin{gathered} {\left( {x + 1} \right)^6} = {}^6{C_0}{x^6} + {}^6{C_1}{x^5} + {}^6{C_2}{x^4} + {}^6{C_3}{x^3} + {}^6{C_4}{x^2} + {}^6{C_5}x + {}^6{C_6} \hfill \\ {\left( {x - 1} \right)^6} = {}^6{C_0}{x^6} - {}^6{C_1}{x^5} + {}^6{C_2}{x^4} - {}^6{C_3}{x^3} + {}^6{C_4}{x^2} - {}^6{C_5}x + {}^6{C_6} \hfill \\ \end{gathered}$

Therefore,

$\begin{gathered} {\left( {x + 1} \right)^6} + {\left( {x - 1} \right)^6} = {}^6{C_0}{x^6} + {}^6{C_1}{x^5} + {}^6{C_2}{x^4} + {}^6{C_3}{x^3} + {}^6{C_4}{x^2} + {}^6{C_5}x + {}^6{C_6} \\ + \left[ {{}^6{C_0}{x^6} - {}^6{C_1}{x^5} + {}^6{C_2}{x^4} - {}^6{C_3}{x^3} + {}^6{C_4}{x^2} - {}^6{C_5}x + {}^6{C_6}} \right] \\ = 2\left( {{}^6{C_0}{x^6} + {}^6{C_2}{x^4} + {}^6{C_4}{x^2} + {}^6{C_6}} \right) \\ = 2\left( {{x^6} + 15{x^4} + 15{x^2} + 1} \right) \\ \end{gathered}$

By putting $x = \sqrt 2$, we obtain

$\begin{gathered} {\left( {\sqrt 2 + 1} \right)^6} + {\left( {\sqrt 2 - 1} \right)^6} = 2\left[ {{{\left( {\sqrt 2 } \right)}^6} + 15{{\left( {\sqrt 2 } \right)}^4} + 15{{\left( {\sqrt 2 } \right)}^2} + 1} \right] \\ = 2\left[ {8 + 15 \cdot 4 + 15 \cdot 2 + 1} \right] \\ = 2\left[ {8 + 60 + 30 + 1} \right] \\ = 2 \times 99 \\ = 198 \\ \end{gathered}$

13. Show that ${9^{n + 1}} - 8n - 9$ is divisible by 64, whenever n is a positive integer.

Ans. In order to show that ${9^{n + 1}} - 8n - 9$ is divisible by 64, it has to be prove that, ${9^{n + 1}} - 8n - 9 = 64k$, where k is some natural number.

By Binomial Theorem,

${\left( {1 + a} \right)^m} = {}^m{C_0} + {}^m{C_1}a + {}^m{C_2}{a^2} + ... + {}^m{C_m}{a^m}$

For $a = 8$ and $m = n + 1$, we obtain

$\begin{gathered} {\left( {1 + 8} \right)^{n + 1}} = {}^{n + 1}{C_0} + {}^{n + 1}{C_1}\left( 8 \right) + {}^{n + 1}{C_2}{\left( 8 \right)^2} + ... + {}^{n + 1}{C_{n + 1}}{\left( 8 \right)^{n + 1}} \\ {9^{n + 1}} = 1 + \left( {n + 1} \right)\left( 8 \right) + {8^2}\left[ {{}^{n + 1}{C_2} + {}^{n + 1}{C_3} \times 8 + ... + {}^{n + 1}{C_{n + 1}}{{\left( 8 \right)}^{n - 1}}} \right] \\ {9^{n + 1}} = 9 + 8n + 64\left[ {{}^{n + 1}{C_2} + {}^{n + 1}{C_3} \times 8 + ... + {}^{n + 1}{C_{n + 1}}{{\left( 8 \right)}^{n - 1}}} \right] \\ {9^{n + 1}} - 8n - 9 = 64k,{\text{ where }}k = {}^{n + 1}{C_2} + {}^{n + 1}{C_3} \times 8 + ... + {}^{n + 1}{C_{n + 1}}{\left( 8 \right)^{n - 1}}{\text{ is a natural number}} \\ \end{gathered}$

Thus, ${9^{n + 1}} - 8n - 9$ is divisible by 64, whenever n is a positive integer.

14. Prove that $\sum\limits_{r = 0}^n {{3^r}{}^n{C_r}} = {4^n}$.

Ans. By Binomial Theorem,

$\sum\limits_{r = 0}^n {{}^n{C_r}{a^{n - r}}{b^r}} = {\left( {a + b} \right)^n}$

By putting $b = 3$ and $a = 1$ in the above equation, we obtain

$\begin{gathered} \sum\limits_{r = 0}^n {{}^n{C_r}{{\left( 1 \right)}^{n - r}}{{\left( 3 \right)}^r}} = {\left( {1 + 3} \right)^n} \\ \sum\limits_{r = 0}^n {{3^r}{}^n{C_r}} = {4^n} \\ \end{gathered}$

Hence proved.

### Miscellaneous Exercise:

1. If a and b are distinct integers, prove that a-b is a factor of ${a^n} - {b^n}$, whenever n is a positive integer

(hint: ${a^n} = {(a - b + b)^n}$)

Ans:To prove to prove that(a-b) is a factor of (${a^n} - {b^n}$), it must be proved that ${a^n} - {b^n}$= $k(a - b)$, where k is some natural number It can be written that, $a = a - b + b$

${((a - b) + b)^n}{ = ^n}{C_0}{(a - b)^n}{ + ^n}{C_1}{(a - b)^{n - 1}}b{ + ^n}{C_2}{(a - b)^{n - 2}}{b^2} + \ldots { + ^n}{C_{n - 1}}(a - b){b^{n - 1}}{ + ^n}{C_n}{b^n}$

=${(a - b)^n}{ + ^n}{C_1}{(a - b)^{n - 1}}b{ + ^n}{C_2}{(a - b)^{n - 2}}{b^2} + \ldots { + ^n}{C_{n - 1}}(a - b){b^{n - 1}} + {b^n}$

=${a^n} - {b^n} = (a - b)$$[{(a - b)^{n - 1}}{ + ^n}{C_1}{(a - b)^{n - 2}}b{ + ^n}{C_2}{(a - b)^{n - 3}}{b^2} + \ldots { + ^n}{C_{n - 1}}{b^{n - 1}}]$

$\Rightarrow {a^n} - {b^n} = k(a - b)$

Where k =$[{(a - b)^{n - 1}}{ + ^n}{C_1}{(a - b)^{n - 2}}b{ + ^n}{C_2}{(a - b)^{n - 3}}{b^2} + \ldots { + ^n}{C_{n - 1}}{b^{n - 1}}]$ is a natural number this shows that $(a - b)$is a factor of $({a^n} - {b^n})$,

Where n is a positive integer.

2. Evaluate $\left( {\sqrt 3 } \right. + {\left. {\sqrt 2 } \right)^6} - \left( {\sqrt 3 } \right. - {\left. {\sqrt 2 } \right)^6}$

Ans: Firstly, the expression

${\left( {a + b} \right)^6} - {\left( {a - b} \right)^6}$ is simplified by using Binomial Theorem. This can be done as

${\left( {a + b} \right)^6}$=$^6{C_0}{(a)^6}{ + ^6}{C_1}{(a)^5}b{ + ^6}{C_2}{(a)^4}{b^2}{ + ^6}{C_3}{(a)^3}{b^3}{ + ^6}{C_4}{(a)^2}{b^4}{ + ^6}{C_5}(a){b^5}{ + ^6}{C_6}{b^6}$

=${(a)^6} + 6{(a)^5}b + 15{(a)^4}{b^2} + 20{(a)^3}{b^3} + 15{(a)^2}{b^4} + 6a{b^5} + {b^6}$

Putting a=$\sqrt 3 \,and\,$b=$\sqrt 2$, we obtain

$\left( {\sqrt 3 } \right. + {\left. {\sqrt 2 } \right)^6} - \left( {\sqrt 3 } \right. - {\left. {\sqrt 2 } \right)^6}$

$= 2\left( {6 \cdot {{(\sqrt 3 )}^5}(\sqrt 2 ) + 20 \cdot {{(\sqrt 3 )}^3}{{(\sqrt 2 )}^3} + 6 \cdot (\sqrt 3 ){{(\sqrt 2 )}^5}} \right)$

=$2 \times 198\sqrt 6$

$= 396\sqrt 6$

3. Find the values of${\left( {{a^2} + \sqrt {{a^2} - 1} } \right)^4} + {\left( {{a^2} - \sqrt {{a^2} - 1} } \right)^4}$

Ans: Firstly, the expression is simplified by using the Binomial Theorem.

${\left( {x + y} \right)^4} + {\left( {x - y} \right)^4}$

This can be done as

${\left( {x + y} \right)^4}$=$^4{C_0}{(x)^4}{ + ^4}{C_1}{(x)^3}y{ + ^4}{C_2}{(x)^2}{y^2}{ + ^4}{C_3}x\,{y^3}{ + ^4}{C_4}{y^4}$

=${(x)^4} + 4{(x)^3}y + 6{(x)^2}{y^2} + 4x\,{y^3} + {y^4}$

${\left( {x - y} \right)^4}$=$^4{C_0}{(x)^4}{ - ^4}{C_1}{(x)^3}y{ - ^4}{C_2}{(x)^2}{y^2}{ - ^4}{C_3}x\,{y^3}{ - ^4}{C_4}{y^4}$

=${(x)^4} - 4{(x)^3}y - 6{(x)^2}{y^2} - 4x\,{y^3} - {y^4}$

Putting x=${a^2}$ and $y = \sqrt {{a^2} - 1} ,$ We obtain

${\left( {{a^2} + \sqrt {{a^2} - 1} } \right)^4} + {\left( {{a^2} - \sqrt {{a^2} - 1} } \right)^4}$

$= 2\left[ {{{\left( {{a^2}} \right)}^4} + 6{{\left( {{a^2}} \right)}^2}{{\left( {\sqrt {{a^2} - 1} } \right)}^2} + {{\left( {\sqrt {{a^2} - 1} } \right)}^4}} \right]$

$= 2\left[ {\left( {{a^8}} \right) + 6\left( {{a^4}} \right)\left( {{a^2} - 1} \right) + {{\left( {{a^2} - 1} \right)}^2}} \right]$

$= 2\left[ {{a^8} + 6{a^6} - 6{a^4} + {a^4} - 2{a^2} + 1} \right]$

$= 2\left[ {{a^8} + 6{a^6} - 5{a^4} - 2{a^2} + 1} \right]$

$= 2{a^8} + 12{a^6} - 10{a^4} - 4{a^2} + 2$

4. Find an approximation of ${\left( {0.99} \right)^5}$using the first three terms of its expansion.

Ans: $0.99\, = 1 - 0.01$

${\left( {0.99} \right)^5} = {\left( {1 - 0.01} \right)^5}$

$^5{C_0}{(1)^5}{ - ^5}{C_1}{(1)^4}\left( {0.01} \right){ - ^5}{C_2}{(1)^3}{\left( {0.01} \right)^2}$

(Approximately)

$= 1 - 0.05 + 0.001$

$= 1.001 - 0.05$

=$= 0.951$

Thus, the value of ${\left( {0.99} \right)^5}$is approximately 0.951

5. Expand using Binomial Theorem ${\left( {1 + \dfrac{x}{2} - \dfrac{2}{x}} \right)^4},\,x \ne 0$

Ans: ${\left( {1 + \dfrac{x}{2} - \dfrac{2}{x}} \right)^4}$

${ = ^n}{C_0}\left( {1 + {{\dfrac{x}{2}}^4}} \right){ - ^n}{C_1}{\left( {1 + {{\dfrac{x}{2}}^4}} \right)^3}\left( {\dfrac{2}{x}} \right) - {\,^n}{C_2}{\left( {1 + {{\dfrac{x}{2}}^4}} \right)^2}{\left( {\dfrac{2}{x}} \right)^2}{ - ^n}{C_3}\left( {1 + {{\dfrac{x}{2}}^4}} \right){\left( {\dfrac{2}{x}} \right)^3}{ - ^n}{C_4}{\left( {\dfrac{2}{x}} \right)^4}$

$= \left( {1 + {{\dfrac{x}{2}}^4}} \right) - 4{\left( {1 + {{\dfrac{x}{2}}^4}} \right)^3}\left( {\dfrac{2}{x}} \right) + \,6\left( {1 + x + {{\dfrac{x}{4}}^2}} \right)\left( {\dfrac{4}{{{x^2}}}} \right) - 4\left( {1 + \dfrac{x}{2}} \right)\left( {\dfrac{8}{{{x^3}}}} \right) + \left( {\dfrac{{16}}{{{x^4}}}} \right)$

$= \left( {1 + {{\dfrac{x}{2}}^4}} \right) - {\left( {1 + {{\dfrac{x}{2}}^4}} \right)^3}\left( {\dfrac{8}{x}} \right) + \,\left( {\dfrac{8}{{{x^2}}}} \right) + \dfrac{{24}}{x} + 6 - \left( {\dfrac{{32}}{{{x^3}}}} \right) + \left( {\dfrac{{16}}{{{x^4}}}} \right)$…..(1)

Again, by using the Binomial Theorem, we obtain

${\left( {1 + \dfrac{x}{2}} \right)^4}{ = ^4}{C_0}{(1)^4}{ + ^4}{C_1}{(1)^3}\left( {\dfrac{x}{2}} \right){ + ^4}{C_2}{(1)^2}{\left( {\dfrac{x}{2}} \right)^2}{ + ^4}{C_3}\,{\left( {\dfrac{x}{2}} \right)^3}{ + ^4}{C_4}{\left( {\dfrac{x}{2}} \right)^4}$

$= 1 + 4 \times \dfrac{x}{2} + 6 \times \dfrac{{{x^4}}}{4} + 4 \times \dfrac{{{x^3}}}{8} + \dfrac{{{x^3}}}{{16}}$

$= 1 + 2x + \dfrac{{3{x^2}}}{2} + \dfrac{{{x^3}}}{2} + \dfrac{{{x^4}}}{{16}}$…..(2)

${\left( {1 + \dfrac{x}{2}} \right)^3}{ = ^3}{C_0}{(1)^3}{ + ^3}{C_1}{(1)^2}\left( {\dfrac{x}{2}} \right){ + ^3}{C_2}(1){\left( {\dfrac{x}{2}} \right)^2}{ + ^3}{C_3}\,{\left( {\dfrac{x}{2}} \right)^3}$

$= \,1 + \dfrac{{3x}}{2} + \dfrac{{3{x^2}}}{4} + \dfrac{{{x^3}}}{8} + \dfrac{{{x^3}}}{8}$…… (3)

From (1), (2), and (3) we obtain

${\left( {\left( {1 + \dfrac{x}{2}} \right) - \dfrac{2}{x}} \right)^4}$

$= 1 + 2x + \dfrac{{3{x^2}}}{2} + \dfrac{{{x^3}}}{2} + \dfrac{{{x^4}}}{{16}} - \left( {\dfrac{8}{x}} \right)\left( {1 + \dfrac{{3x}}{2} + \dfrac{{3{x^2}}}{4} + \dfrac{{{x^3}}}{8}} \right) + \dfrac{8}{{{x^2}}} + \dfrac{{24}}{x} + 6 - \dfrac{{32}}{{{x^3}}} + \dfrac{{16}}{{{x^4}}}$

$= 1 + 2x + \dfrac{{3{x^2}}}{2} + \dfrac{{{x^3}}}{2} + \dfrac{{{x^4}}}{{16}} - \dfrac{8}{x} - 12 - 6x - {x^2} - \dfrac{8}{{{x^2}}} + \dfrac{{24}}{x} + 6 - \dfrac{{32}}{{{x^3}}} + \dfrac{{16}}{{{x^4}}}$

$= \dfrac{{16}}{x} + \dfrac{8}{{{x^2}}} - \dfrac{{32}}{{{x^3}}} + \dfrac{{16}}{{{x^4}}} - 4x + \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{2} + \dfrac{{{x^4}}}{{16}} - 5$.

6. Find the expansion of ${\left( {3{x^2} - 2ax + 3{a^2}} \right)^3}$using binomial theorem.

Ans: Using the Binomial Theorem, the given expression

${\left( {3{x^2} - 2ax + 3{a^2}} \right)^3}$Can be expanded as

${\left( {3{x^2} - 2ax + 3{a^2}} \right)^3}$

${ = ^3}{C_0}{\left( {3{x^2} - 2ax} \right)^3}{ - ^3}{C_1}{\left( {3{x^2} - 2ax} \right)^2}\left( {3{a^2}} \right){ + ^3}{C_2}\left( {3{x^2} - 2ax} \right){\left( {3{a^2}} \right)^2}{ - ^3}{C_3}{\left( {3{a^2}} \right)^3}$

$= {\left( {3{x^2} - 2ax} \right)^3} + 3\left( {9{x^4} - 12a{x^3} + 4{a^2}{x^2}} \right)\left( {3{a^2}} \right) + 3\left( {3{x^2} - 2ax} \right)\left( {9{a^4}} \right) + \left( {2{a^6}} \right)$

$= {\left( {3{x^2} - 2ax} \right)^3} + 81{a^2}{x^4} - 108{a^3}{x^3} + 36{a^4}{x^2} + 81{a^4}{x^2} - 54{a^5}x + 27{a^6}$

$= {\left( {3{x^2} - 2ax} \right)^3} + 81{a^2}{x^4} - 108{a^3}{x^3} + 117{a^4}{x^2} - 54{a^5}x + 27{a^6}$…. (1)

Again, by using the Binomial Theorem, we obtain

${\left( {3{x^2} - 2ax} \right)^3}$

${ = ^3}{C_0}{\left( {3{x^2}} \right)^3}{ - ^3}{C_1}{\left( {3{x^2}} \right)^2}\left( {2ax} \right){ + ^3}{C_2}\left( {3{x^2}} \right){\left( {2ax} \right)^2}{ - ^3}{C_3}{\left( {2ax} \right)^3}$

$= \left( {27{x^6}} \right) - 3\left( {9{x^4}} \right)\left( {2ax} \right) + 3\left( {3{x^2}} \right)\left( {4{a^2}{x^2}} \right) - 8{a^3}{x^3}$

$= 27{x^6} - 54a{x^5} + 36{a^2}{x^4} - 8{a^3}{x^3}$……… (2)

From (1) and (2), we obtain

${\left( {3{x^2} - 2ax + 3{a^2}} \right)^3}$

$= 27{x^6} - 54a{x^5} + 36{a^2}{x^4} - 8{a^3}{x^3} + 81{a^2}{x^4} - 108{a^3}{x^3} + 117{a^4}{x^2} - 54{a^5}x + 27{a^6}$

$= 27{x^6} - 54a{x^5} + 117{a^2}{x^4} - 116{a^3}{x^3} + 117{a^4}{x^2} - 54{a^5}x + 27{a^6}$.

## Overview of Deleted Syllabus for CBSE Class 11 Maths Binomial Theorem

 Chapter Dropped Topics Binomial Theorem General Middle Terms Question: 1–3, and 8 (Miscellaneous Exercise) The last two points in the Summary

## Class 11 Maths Chapter 7: Exercises Breakdown

 Exercise Number of Questions Exercise 7.1 14 Questions and Solutions Miscellaneous Exercise 6 Questions and Solutions

## Conclusion

Binomial Theorem Class 11 NCERT Solutions is essential for mastering the expansion of binomial expressions and understanding binomial coefficients. It is important to focus on the binomial theorem formula, Pascal's Triangle, and the general and middle terms of the expansion. Practicing these concepts through Vedantu's solutions will help reinforce your understanding. In the previous year's CBSE exams, around 3 to 4 questions were asked from this chapter. By thoroughly studying and practicing, you can enhance your problem-solving skills and perform well in your exams.

## Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions Class 11 Maths Chapter 7 Binomial Theorem

1. What is the Binomial Theorem?

The binomial theorem is defined as the process of algebraically expanding the power of sums of two or more binomials. Coefficients of binomial terms in the process of expansion are referred to as binomial coefficients. The introductory parts of these chapters consist of proper definitions of different aspects of the binomial theorem.

With the Binomial Theorem Class 11 NCERT Solutions PDF, students can now study with ease and be updated with all information that might appear in their examinations. Learning the concepts of the Binomial Theorem will be easier with the NCERT Solutions available on Vedantu.

2. How will the NCERT Solutions Class 11 Maths Chapter 7 Binomial Theorem help in understanding the concepts of the Binomial Theorem?

To learn about the expansion procedure, refer to the examples available in binomial theorem class 11 NCERT solutions. These examples have been solved in a step-by-step format that will help students understand the concepts better.

Students are required to be well-versed in solving these equations if they wish to score well in examinations. There are also exercises in ch 7 maths class 11 that are given in this segment that students can solve on their own. This will allow them to practice what they have learned and clear any doubt that they might have related to the binomial theorem.

3. What are the properties of positive integers in the Binomial Theorem?

There are more than 10 properties that are listed under positive integers that students can learn when studying binomial theorem class 11 NCERT solutions. Students are required to study these properties to understand the basic concept of solving such equations.

Examination papers may target aspects of chapters that seem simple to students but may be tricky to solve, which is why students will need to go through these binomial theorem class 11 NCERT solutions thoroughly if they wish to score well in their upcoming examinations.

4. Explain the concept of the Binomial Theorem covered in binomial theorem class 11.

The Binomial theorem states, for positive integer n, whenever you add any two numbers, say a and b, the result raised to the power of n can also be written as the sum of (n+10 terms). The coefficients involved are expressed as binomial coefficients. The NCERT solutions of Class 11 Maths Chapter 7 can be accessed on the Vedantu website and the app. Practice all of these judiciously if you want to score well in them and in the other topics related to them.

5. What Chapter is Binomial Theorem Class 11?

The Binomial Theorem is Chapter 7 of the NCERT Mathematics book. It explains in detail the Binomial Theorem and also provides the necessary exercises for a better understanding of the concepts by the students. The solutions of ch 7 maths class 11 can easily be found on the Vedantu site (vedantu.com). The students are advised to go through all topics to efficiently grasp the content and score well in their exams.

6. Which is the best Solution book for NCERT Class 11 Chapter 7 Maths?

The best solution book for Chapter 7-Binomial Theorem is available on Vedantu. NCERT Class 11 Maths Chapter 7 Solutions may be obtained by visiting the Vedantu website. Aside from that, you may access a range of modules that will help you achieve excellent grades in math exams. The link to the solution to the exercise is provided below. Visit the page NCERT Solutions for Class 11 Chapter 7 to download the PDF file free of cost.

7. How can I master Class 11 Maths Chapter 7?

There is no easy or singular motto for performing well or ranking high in any topic. To do well, one must constantly be diligent about the core ideas. It is important to practice as many questions as possible thoroughly so that they have a good understanding of the concepts. Proper practice of Chapter 7 class 11 maths is critical for getting high grades and gaining deeper knowledge. Those interested in accessing NCERT solutions, revision notes and important questions of this chapter, visit the Vedantu website or download the Vedantu app.

8. Do I need to practice all the questions provided in Class 11 Maths Chapter 7 NCERT Solutions?

Yes, it is critical to practice and answer all questions since they cover a variety of subjects and concepts which will give you a good understanding of the kind of questions that might be set from those areas. These questions also help you learn how different questions from the same topic may be set. Each exercise should be thoroughly practised. You can discover modules on the Vedantu site or in the Vedantu app that are relevant to this topic or other topics in this Chapter 7 Class 11 maths.

9. What is Pascal's Triangle in binomial theorem class 11 solutions pdf?

Pascal's Triangle is a triangular array of numbers. Each number is the sum of the two numbers directly above it. This pattern is used to find binomial coefficients. It is helpful in expanding binomial theorem class 11 solutions pdf.