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NCERT Solutions

Class 11

Maths

Chapter 7 Binomial Theorem Exercise 7.1

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

Q: 3. When solving an expansion of the form (a - b)ⁿ, what is the correct way to manage the alternating signs in the solution?

To avoid sign errors when expanding (a - b)ⁿ, treat it as [a + (-b)]ⁿ. The most reliable method is to use the general term formula Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ᵣ (-b)ᵣ. The term (-b)ᵣ will be positive if 'r' is even and negative if 'r' is odd. This automatically generates the alternating signs (+, -, +, -, ...) in the expansion, ensuring accuracy.

Q: 5. Why is the general term formula so important for finding the middle term(s) in a binomial expansion for Class 11 Maths?

The general term formula, Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ᵣ bᵣ, is fundamental because it allows you to calculate any term without writing the full expansion. For finding the middle term:If 'n' is even, there is one middle term at the position (n/2 + 1). You find 'r' by setting r = n/2.If 'n' is odd, there are two middle terms at positions ((n+1)/2) and ((n+1)/2 + 1).The formula provides a direct method to pinpoint and calculate these specific terms, which is a standard question type in NCERT Chapter 7.

Q: 6. How can Pascal's Triangle be used as a quick verification method for binomial coefficients in the initial exercises of Chapter 7?

While the ⁿCᵣ formula is the formal method, Pascal's Triangle offers a visual and quick way to check your coefficients for small values of 'n'. Each row of the triangle corresponds to the coefficients of (a+b)ⁿ. For example, the row '1 4 6 4 1' provides the coefficients for n=4. This is a useful tool for verifying answers in early exercises to build confidence in the ⁿCᵣ calculation method.

Q: 8. What key concepts from Chapter 6, Permutations and Combinations, are required to correctly solve the problems in Binomial Theorem?

A strong foundation in Permutations and Combinations is essential before starting Binomial Theorem. The most critical concept is the calculation and understanding of combinations (ⁿCᵣ). This includes:The formula ⁿCᵣ = n! / [r!(n-r)!].Properties like ⁿCᵣ = ⁿCₙ₋ᵣ.Understanding that ⁿCᵣ represents the number of ways to choose 'r' items from 'n', which is the theoretical basis for the binomial coefficients.

Download Free PDF of Binomial Theorem Exercise 7.1 Solutions for Class 11 Maths

Struggling with binomial expansion in Class 11 Maths? Chapter 7, Binomial Theorem, is foundational for mastering algebraic expansion and is weighted for board exams as part of the Algebra unit. Here, you’ll tackle Exercise 7.1 using the general and middle term concepts from your latest CBSE syllabus.

Table of Content

Students often search for “exercise 7.1 class 11” when they want stepwise NCERT answers that make every coefficient calculation clear. By following each resolved example, you’ll learn to use algebraic expansion steps confidently, recognize coefficient pattern formulas, and avoid common mistakes around properties of binomial coefficients.

Every detailed solution is checked by Vedantu’s subject experts for accuracy and clarity. With these NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.1, you can build a strong foundation for exam success—one step at a time.

Access NCERT Solutions for Maths Class 11 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, 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

Exercise	Number of Questions
Miscellaneous Exercise	6 Questions & Solutions

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.

CBSE Class 11 Maths Chapter 7 Other Study Materials

S. No	Important Links for Chapter 7 Binomial Theorem
1	Class 11 Binomial Theorem Important Questions
2	Class 11 Binomial Theorem Revision Notes
3	Class 11 Binomial Theorem Important Formulas
4	Class 11 Binomial Theorem NCERT Exemplar Solution
5	Class 11 Binomial Theorem RD Sharma Solutions
6	Class 11 Binomial Theorem RS Aggarwal Solutions

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.

S. No	NCERT Solutions Class 11 Maths All Chapters
1	Chapter 1 - Sets Solutions
2	Chapter 2 - Relations and Functions Solutions
3	Chapter 3 - Trigonometric Functions Solutions
4	Chapter 4 - Complex Numbers and Quadratic Equations Solutions
5	Chapter 5 - Linear Inequalities Solutions
6	Chapter 6 - Permutations and Combinations Solutions
7	Chapter 7 - Binomial Theorem Solutions
8	Chapter 8 - Sequences and Series Solutions
9	Chapter 9 - Straight Lines Solutions
10	Chapter 10 - Conic Sections Solutions
11	Chapter 11 - Introduction to Three Dimensional Geometry Solutions
12	Chapter 12 - Limits and Derivatives Solutions
13	Chapter 13 - Statistics Solutions
14	Chapter 14 - Probability Solutions

Important Related Links for CBSE Class 11 Maths

S.No.	Important Study Material for Maths Class 11
1	CBSE Class 11 Maths NCERT Books
2	CBSE Class 11 Maths Revision Notes
3	CBSE Class 11 Maths Important Questions
4	CBSE Class 11 Maths NCERT Solutions
5	CBSE Class 11 Maths NCERT Exemplar Solution
6	CBSE Class 11 Maths RS Aggarwal Solutions
7	CBSE Class 11 Maths RD Sharma Solutions
8	CBSE Class 11 Maths Formulas
9	CBSE Class 11 Maths Syllabus
10	CBSE Class 11 Maths Sample Papers

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

1. How do I correctly apply the Binomial Theorem to expand an expression like (x + y)ⁿ as per the NCERT solutions for Class 11 Maths Chapter 7?

To correctly expand any binomial expression (a + b)ⁿ according to the CBSE 2025-26 syllabus, you must follow a step-by-step method using the main formula:
(a + b)ⁿ = Σⁿᵣ₌₀ ⁿCᵣ aⁿ⁻ᵣ bᵣ
The standard procedure is:

Identify 'a', 'b', and 'n' in the given expression.
Calculate the binomial coefficients ⁿCᵣ for each term, from r=0 to r=n.
Systematically write out each term by substituting the values of a, b, and the coefficients.
Simplify the powers and numerical values for the final answer.

2. What is the step-by-step method to find the general term, Tᵣ₊₁, in a binomial expansion for solving Chapter 7 problems?

The general term formula is crucial for solving most NCERT questions in this chapter. The correct formula is Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ᵣ bᵣ. To apply it:

Identify the components a, b, and n from the expression (a + b)ⁿ.
Write down the formula for the general term.
Substitute the values of a, b, and n into the formula.
Simplify the expression to get a formula for any term based on the value of 'r'. This method is essential for finding specific terms, coefficients, or terms independent of a variable.

3. When solving an expansion of the form (a - b)ⁿ, what is the correct way to manage the alternating signs in the solution?

To avoid sign errors when expanding (a - b)ⁿ, treat it as [a + (-b)]ⁿ. The most reliable method is to use the general term formula Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ᵣ (-b)ᵣ. The term (-b)ᵣ will be positive if 'r' is even and negative if 'r' is odd. This automatically generates the alternating signs (+, -, +, -, ...) in the expansion, ensuring accuracy.

4. What are the most common calculation mistakes to avoid when finding binomial coefficients (ⁿCᵣ) in the NCERT exercises?

Common mistakes in calculating ⁿCᵣ often lead to incorrect solutions. To avoid them:

Factorial Calculation: Double-check the factorial calculations in the formula ⁿCᵣ = n! / [r! * (n-r)!]. A small error here affects the entire term.
Symmetry Property: Remember that ⁿCᵣ = ⁿCₙ₋ᵣ. Using this can simplify calculations, for example, ¹⁰C₈ is easier to calculate as ¹⁰C₂.
Special Cases: Always remember that ⁿC₀ = 1 and ⁿCₙ = 1. This helps speed up the expansion of the first and last terms.

5. Why is the general term formula so important for finding the middle term(s) in a binomial expansion for Class 11 Maths?

The general term formula, Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ᵣ bᵣ, is fundamental because it allows you to calculate any term without writing the full expansion. For finding the middle term:

If 'n' is even, there is one middle term at the position (n/2 + 1). You find 'r' by setting r = n/2.
If 'n' is odd, there are two middle terms at positions ((n+1)/2) and ((n+1)/2 + 1).

The formula provides a direct method to pinpoint and calculate these specific terms, which is a standard question type in NCERT Chapter 7.

6. How can Pascal's Triangle be used as a quick verification method for binomial coefficients in the initial exercises of Chapter 7?

While the ⁿCᵣ formula is the formal method, Pascal's Triangle offers a visual and quick way to check your coefficients for small values of 'n'. Each row of the triangle corresponds to the coefficients of (a+b)ⁿ. For example, the row '1 4 6 4 1' provides the coefficients for n=4. This is a useful tool for verifying answers in early exercises to build confidence in the ⁿCᵣ calculation method.

7. For problems that require finding a term independent of a variable (like x), what is the standard procedure using the Binomial Theorem?

The correct procedure for finding a term independent of x is as follows:

Write the general term Tᵣ₊₁ of the expansion.
Collect all powers of x and simplify them into a single term, like xᵏ, where 'k' is an expression involving 'r'.
To find the term independent of x, set the exponent k to zero (k = 0).
Solve the resulting equation to find the value of 'r'.
Substitute this value of 'r' back into the general term formula to get the required constant term.

8. What key concepts from Chapter 6, Permutations and Combinations, are required to correctly solve the problems in Binomial Theorem?

A strong foundation in Permutations and Combinations is essential before starting Binomial Theorem. The most critical concept is the calculation and understanding of combinations (ⁿCᵣ). This includes:

The formula ⁿCᵣ = n! / [r!(n-r)!].
Properties like ⁿCᵣ = ⁿCₙ₋ᵣ.
Understanding that ⁿCᵣ represents the number of ways to choose 'r' items from 'n', which is the theoretical basis for the binomial coefficients.

9. In problems asking to find the value of expressions like (101)⁴, how should the Binomial Theorem be applied to simplify the calculation?

To solve such problems, you should first break the number into a simple binomial form. For example:

Rewrite (101)⁴ as (100 + 1)⁴.
Here, a=100, b=1, and n=4.
Apply the Binomial Theorem: (100 + 1)⁴ = ⁴C₀(100)⁴ + ⁴C₁(100)³(1)¹ + ⁴C₂(100)²(1)² + ...
This method simplifies the problem into calculations with powers of 10, which are much easier to handle than direct multiplication.

10. How do you determine which number is larger in a comparison problem, such as (1.01)¹⁰⁰⁰⁰⁰⁰ or 10,000, using the binomial expansion method?

To solve comparison problems using the Binomial Theorem, you should expand the expression and analyse its initial terms. For (1.01)¹⁰⁰⁰⁰⁰⁰:

Rewrite it as (1 + 0.01)¹⁰⁰⁰⁰⁰⁰.
Expand using the theorem: (1 + 0.01)¹⁰⁰⁰⁰⁰⁰ = ¹⁰⁰⁰⁰⁰⁰C₀(1)¹⁰⁰⁰⁰⁰⁰ + ¹⁰⁰⁰⁰⁰⁰C₁(1)⁹⁹⁹⁹⁹⁹(0.01)¹ + ...
Calculate the first two terms: The first term is 1. The second term is 1,000,000 × 0.01 = 10,000.
The sum of just the first two terms is 1 + 10,000 = 10,001.
Since the expansion contains many other positive terms, the total value is clearly greater than 10,001, and therefore much larger than 10,000.