CBSE Class 11 Maths Chapter 7 Binomial Theorem – NCERT Solutions 2025-26

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, 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, 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, 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, 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 Binomial Theorem, the first few terms of ${\left( {1.1} \right)^{10000}}$ 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.

NCERT Solution Class 11 Maths of Chapter 7 All Exercises

Conclusion

In class 11 Binomial Theorem Exercise 7.1, we explored the fundamentals of the Binomial Theorem and its applications. We learned how to expand expressions raised to a power using the Binomial Theorem for positive integral indices. The exercise also introduced us to Pascal’s Triangle, which provides an easy way to find binomial coefficients. By practicing these problems, you have gained a deeper understanding of how to use the Binomial Theorem to simplify complex algebraic expressions. This knowledge is essential for solving higher-level mathematics problems and will be useful in many areas of study.

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