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NCERT Solutions for Class 11 Maths Chapter 7 - Binomial Theorem Exercise 7.1

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NCERT Solutions for Maths Class 11 Chapter 7 Binomial Theorem Exercise 7.1 - FREE PDF Download

NCERT Solution for Class 11 Maths Binomial Theorem Exercise 7.1 Students will learn about the Binomial Theorem, a powerful algebra tool. The Binomial Theorem helps us expand expressions that are raised to a power, like $(a+b)^n$

Table of Content

1. NCERT Solutions for Maths Class 11 Chapter 7 Binomial Theorem Exercise 7.1 - FREE PDF Download

2. Glance on NCERT Solutions Maths Chapter 7 Exercise 7.1 Class 11 | Vedantu

3. Formulas Used in Class 11 Maths Ch 7 Ex 7.1

4. Access NCERT Solutions for Maths Class 11 Chapter 7 - Binomial Theorem

4.1

Exercise 7.1

5. CBSE Class 11 Maths Chapter 7 Other Study Materials

6. Chapter-Specific NCERT Solutions for Class 11 Maths

FAQs

Exercise 7.1 focuses on understanding the basics of the Binomial Theorem. It will guide you through expanding binomials and finding coefficients of terms in the expansion. This key concept will help you solve many algebraic problems more easily. Access the Class 11 Maths Syllabus here.

Glance on NCERT Solutions Maths Chapter 7 Exercise 7.1 Class 11 | Vedantu

In NCERT Solutions for Class 11 Maths, the Binomial Theorem Exercise 7.1 covers topics such as the Binomial Theorem for Positive Integral Indices, Pascal’s Triangle, and the Binomial theorem for any positive integer n.
The Binomial Theorem for Positive Integral Indices provides a formula for expanding expressions of the form $(a + b)^n$, where n is a positive whole number (1, 2, 3, ...).
The formula helps you determine the coefficients (numbers multiplying each term) and exponents (powers of a and b) in the expanded expression.
Class 11 Maths Ch 7 Ex 7.1 uses this formula to expand binomials raised to positive integer powers.
Although not directly related to Exercise 7.1, Pascal's Triangle is a triangular arrangement of numbers where each internal number is the sum of the two numbers above it in the triangle.
The connection between Pascal's Triangle and the Binomial Theorem comes later. The coefficients in the Binomial Theorem expansion (mentioned in point 1) can be derived from Pascal's Triangle.
The binomial theorem for any positive integer n is a more advanced concept that extends the Binomial Theorem beyond positive integer powers. It allows expanding expressions like $(a + b)^n$ even when n is a fraction or a negative number (with certain restrictions).
Ex 7.1 Class 11 contains 14 Questions and Solutions.

Formulas Used in Class 11 Maths Ch 7 Ex 7.1

Binomial theorem for any positive integer n:
$(a+b)^{n}=^{n}C_{0}a^{n}+^{n}C_{1}a^{n-1}b+^{n}C_{2}a^{(n-2)}b^{2}+\cdots +^{n}C_{(n-1)}ab^{(n-1)}+^{n}C_{n}b^{n}$

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 CBSE Class 11 Maths
1	CBSE Class 11 Maths NCERT Books
2	CBSE Class 11 Maths Important Questions
3	CBSE Class 11 Maths Revision Notes
4	CBSE Class 11 Maths RS Aggarwal Solutions
5	CBSE Class 11 Maths RD Sharma Solutions

FAQs on NCERT Solutions for Class 11 Maths Chapter 7 - Binomial Theorem Exercise 7.1

1. What are some of the useful Binomial expansions?

Some useful binomial expansions are as follows:

(x + y)n + (x−y)n = 2[C0 xn + C2 xn-1 y2 + C4 xn-4 y4 + …]
(x + y)n – (x−y)n = 2[C1 xn-1 y + C3 xn-3 y3 + C5 xn-5 y5 + …]
(1 + x)n = nΣr-0 nCr . xr = [C0 + C1 x + C2 x2 + … Cn xn]
(1+x)n + (1 − x)n = 2[C0 + C2 x2+C4 x4 + …]
(1+x)n − (1−x)n = 2[C1 x + C3 x3 + C5 x5 + …]
The number of terms in the expansion of (x + a)n + (x−a)n is (n+2)/2 if “n” is even or (n+1)/2 if “n” is odd.
The number of terms in the expansion of (x + a)n − (x−a)n is (n/2) if “n” is even or (n+1)/2 if “n” is odd.

2. What are the Properties of Binomial Coefficients?

The various Properties of Binomial Coefficients are as follows:

C0 + C1 + C2 + … + Cn = 2n
C0 + C2 + C4 + … = C1 + C3 + C5 + … = 2n-1
C0 – C1 + C2 – C3 + … +(−1)n . nCn = 0
nC1 + 2.nC2 + 3.nC3 + … + n.nCn = n.2n-1
C1 − 2C2 + 3C3 − 4C4 + … +(−1)n-1 Cn = 0 for n > 1
C02 + C12 + C22 + …Cn2 = [(2n)!/ (n!)2]

3. What are the few important features of Binomial Expansion that students must know in class 11 binomial theorem Exercise 7.1?

Some important features of Binomial Expansion that students must know are as follows:

The total number of terms in the expansion of (x+y)n is (n+1)
The sum of exponents of x and y is always n.
nC0, nC1, nC2, … .., nCn are called binomial coefficients and also represented by C0, C1, C2, ….., Cn
The binomial coefficients which are equidistant from the beginning and the ending are equal i.e. nC0 = nCn, nC1 = nCn-1, nC2 = nCn-2,….. Etc.

4. What are the various Applications of the Binomial Theorem according to Ex 7.1 Class 11 Maths NCERT Solutions?

The various Applications of Binomial Theorem are as follows:

Finding Remainder using Binomial Theorem
Ex: Find the remainder when 7105 is divided by 25
Finding Digits of a Number
Ex: Find the last two digits of the number (13)10
Relation Between Two Numbers
Ex: Find the larger of 9950 + 10050 and 10150
Divisibility Test
Ex: Show that 119 + 911 is divisible by 10.

5. Which is the best source for Class 11 Maths Chapter 7 Exercise 7.1 for exam preparation?

The best resource for CBSE exam preparation is NCERT textbooks. Students must refer to NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.1 provided by Vedantu as it is one of the best study materials which provides them with a clear understanding of basics easily and simply. Studying these thoroughly before the exams will help the students improve their problem-solving abilities and score good marks. These solutions which are available free of cost on the Vedantu website will also help the students gain a step-by-step understanding of the binomial theorem.

6. Explain the properties of positive integers in the Binomial Theorem.

In the Class 11 Ex 7.1 Binomial Theorem, students will get to learn different properties of positive integers by using the Ex 7.1 Class 11 Maths NCERT Solutions Students must understand and learn these properties as they are important to solving the equations efficiently. In the annual exam, students will be asked questions from easy chapters that are tricky to solve. That is why students are recommended to go through the NCERT Solutions to score good marks.

7. What is the concept of the Binomial Theorem covered in Chapter 7 of NCERT Solutions for Class 11 Maths Exercise 7.1?

The process of expanding the power of sums of two or more binomials algebraically is known as the Binomial Theorem. In the process of expansion, the coefficients of binomial terms that are involved are called binomial coefficients. The chapter’s introduction has definitions of terms that the students have to study as they are important for the exams. Students can now refer to the NCERT Solutions which are available in PDF format to study and be updated about the latest syllabus of the CBSE board.

8. Why should students download NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.1 from Vedantu?

Vedantu provides students with the most accurate NCERT solutions making it easier for the students in their homework as well as exam preparation. These NCERT solutions can be viewed online and can also be downloaded in PDF format for free from the official website of Vedantu. The solutions for this exercise are explained in a stepwise and detailed manner very clearly by the Maths experts. Also, all the solutions are framed keeping in mind the guidelines of the CBSE. These solutions are available at free of cost on Vedantu(vedantu.com) and mobile app.

9. Do I need to practice all the questions from Class 11 Maths Chapter 7 Exercise 7.1?

Yes. Students should practice all the questions from Class 11 Maths Chapter 7 Exercise 7.1. Students can refer to the NCERT solutions provided by Vedantu. This will help the students to understand the concepts clearly. Practicing once or twice will be a quick revision of all the concepts and also students will know if they have difficulty in understanding Class 11 Maths Binomial Theorem Exercise 7.1. Students should use their logical and analytical skills to solve them. It will also improve the student’s accuracy and speed.

10. What is Pascal’s Triangle, and how is it used in the Binomial Theorem explain it as you studied in Ex 7.1 Class 11 Maths NCERT Solutions.

Pascal’s Triangle is a triangular array of numbers where each number is the sum of the two numbers directly above it. The rows of Pascal’s Triangle correspond to the coefficients in the binomial expansion, providing a quick way to find these coefficients.

11. Can the Binomial Theorem be used for any positive integer n?

Yes, the Binomial Theorem applies to any positive integer n. It allows the expansion of (a+b)^n into a series of terms with coefficients given by binomial coefficients. Understand more about this formula in Class 11 Ex 7.1.