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

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

NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.1 Binomial Theorem help students understand the basic concepts of the Binomial Theorem and its application in expanding binomial expressions. This exercise introduces the general form of the binomial expansion, which allows students to expand expressions like

$(a + b)^n$

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

Binomial Theorem

Medium

Using binomial theorem, evaluate the following: (99)⁵

Binomial Theorem

Medium

Find a, if the coefficients of x² and x³ in the expansion of (3 + ax)⁹ are equal.

Binomial Theorem

Medium

Find the middle term in the expansion of {(\frac{x}{3} +9y)}^{10} .

Binomial Theorem

Medium

In the expansion of (1 + a)^m+n, prove that coefficients of a^m and aⁿ are equal.

Binomial Theorem

Difficult

The coefficients of the (r – 1)^th, r^th and (r + 1)^th terms in the expansion of (x + 1)ⁿare in the ratio 1 : 3 : 5. Find n and r.

Binomial Theorem

Difficult

Prove that the coefficient of xⁿ in the expansion of (1 + x)²ⁿ is twice the coefficient of xⁿ in the expansion of (1 + x)^{2n – 1}.

Binomial Theorem

Medium

Find a positive value of m for which the coefficient of x² in the expansion (1 + x)^m is 6.

Binomial Theorem

Medium

Find a, b and n in the expansion of (a + b)ⁿ if the first three terms of the expansion are 729, 7290 and 30375, respectively.

Binomial Theorem

Medium

Find the coefficient of x⁵ in the product (1 + 2x)⁶ (1 – x)⁷ using binomial theorem.

Binomial Theorem

Easy

Write {the general term in the expansion of (x}^{2} - yx)^{12}, x \neq 0 .

Binomial Theorem

Difficult

If a and b are distinct integers, prove that a – b is a factor of aⁿ – bⁿ, whenever n is a positive integer.

Binomial Theorem

Difficult

Evaluate : {(\sqrt{3} + \sqrt{2})}^{6} - {(\sqrt{3} - \sqrt{2})}^{6} .

Binomial Theorem

Difficult

Find the value of {(a^{2} + \sqrt{a^{2} - 1})}^{4} + {(a^{2} - \sqrt{a^{2} - 1})}^{4} .

Binomial Theorem

Easy

Find an approximation of (0.99)⁵ using the first three terms of its expansion.

Binomial Theorem

Difficult

\begin{array}{l} Find n, if the ratio of the fifth term from the beginning to the fifth term from the end \\ in the expansion of {(\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}})}^{n} is \sqrt{6} :1. \end{array}

Binomial Theorem

Medium

Expand using Binomial Theorem {(1+ \frac{x}{2} - \frac{2}{x})}^{4},  x \neq 0.

Binomial Theorem

Medium

Find the middle term in the expansion of {(3 - \frac{x^{3}}{6})}^{7} .

Binomial Theorem

Medium

{Find the 13}^{th} term in the expansion of {(9x - \frac{1}{3 \sqrt{x}})}^{18}, x \neq 0.

Binomial Theorem

Easy

Expand the expression: (1 – 2x)⁵.

Binomial Theorem

Medium

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

Binomial Theorem

Medium

Expand the expression: {(\frac{2}{x} - \frac{x}{2})}^{5} .

Binomial Theorem

Medium

Expand the expression: (2x – 3)⁶.

Binomial Theorem

Difficult

Expand the expression: {(\frac{x}{3} + \frac{1}{x})}^{5} .

Binomial Theorem

Difficult

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

Binomial Theorem

Medium

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

Binomial Theorem

Medium

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

Binomial Theorem

Medium

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

Binomial Theorem

Difficult

Find the 4^th term in the expansion of (x – 2y)¹².

Binomial Theorem

Difficult

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

Binomial Theorem

Difficult

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

Binomial Theorem

Medium

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

Binomial Theorem

Medium

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

Binomial Theorem

Medium

Find the coefficient of x⁵ in (x + 3)⁸.

Binomial Theorem

Medium

Find the coefficient of a⁵b⁷in (a – 2b)¹².

Binomial Theorem

Easy

Write the general term in the expansion of (x² – y)⁶.

Binomial Theorem

Difficult

Find the expansion of (3x² – 2ax + 3a²)³ using binomial theorem.

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

$(a + b)^{n}$ .

Prepared according to the latest CBSE Class 11 Maths syllabus, Exercise 7.1 focuses on understanding and applying the Binomial Theorem for positive integer exponents. The exercise includes problems where students will expand binomial expressions and find specific terms in the expansion.

The solutions are explained in a clear, step-by-step format so students can easily understand how to expand binomial expressions and solve related problems confidently.

1. Expand the expression

$(1 - 2x)^5$

$(1 - 2 x)^{5}$

By using the Binomial Theorem, the expression

$(1 - 2x)^5$

$(1 - 2 x)^{5}$ can be expanded as:

$(1 - 2x)^5 = C_0(1) + C_1(1)(-2x) + C_2(1^2)(-2x)^2 + C_3(1^3)(-2x)^3 + C_4(1^4)(-2x)^4 + C_5(1^5)(-2x)^5$

$(1 - 2 x)^{5} = C_{0} (1) + C_{1} (1) (- 2 x) + C_{2} (1^{2}) (- 2 x)^{2} + C_{3} (1^{3}) (- 2 x)^{3} + C_{4} (1^{4}) (- 2 x)^{4} + C_{5} (1^{5}) (- 2 x)^{5}$

Expanding this:

$= 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5$

$= 1 - 10 x + 40 x^{2} - 80 x^{3} + 80 x^{4} - 32 x^{5}$

2. Expand the expression

$\left(\frac{2}{x} - \frac{x}{2}\right)^5$

$(x 2 - 2 x)^{5}$

By using the Binomial Theorem, the expression

$\left(\frac{2}{x} - \frac{x}{2}\right)^5$

$(x 2 - 2 x)^{5}$ can be expanded as:

$\left(\frac{2}{x} - \frac{x}{2}\right)^5 = C_0\left(\frac{2}{x}\right)^5 + C_1\left(\frac{2}{x}\right)^4\left(-\frac{x}{2}\right) + C_2\left(\frac{2}{x}\right)^3\left(-\frac{x}{2}\right)^2 + \ldots$

$(x 2 - 2 x)^{5} = C_{0} (x 2)^{5} + C_{1} (x 2)^{4} (- 2 x) + C_{2} (x 2)^{3} (- 2 x)^{2} + \dots$

Expanding this:

$= \frac{32}{x^5} - \frac{80}{x^3} + 40x - 10x^2 + 5x^3 + x^5$

$= x ^{5} 32 - x ^{3} 80 + 40 x - 10 x^{2} + 5 x^{3} + x^{5}$

3. Expand the expression

$(2x - 3)^6$

$(2 x - 3)^{6}$

By using the Binomial Theorem, the expression

$(2x - 3)^6$

$(2 x - 3)^{6}$ can be expanded as:

$(2x - 3)^6 = C_0(2x)^6 + C_1(2x)^5(-3) + C_2(2x)^4(-3)^2 + \ldots$

$(2 x - 3)^{6} = C_{0} (2 x)^{6} + C_{1} (2 x)^{5} (- 3) + C_{2} (2 x)^{4} (- 3)^{2} + \dots$

Expanding this:

$= 64x^6 - 576x^5 + 2160x^4 - 4320x^3 + 4860x^2 - 2916x + 729$

$= 64 x^{6} - 576 x^{5} + 2160 x^{4} - 4320 x^{3} + 4860 x^{2} - 2916 x + 729$

4. Expand the expression

$(2x - 3)^6$

$(2 x - 3)^{6}$

Using the Binomial Theorem:

$(2x - 3)^6 = C_0(2x)^6 + C_1(2x)^5(-3) + C_2(2x)^4(-3)^2 + C_3(2x)^3(-3)^3 + C_4(2x)^2(-3)^4 + C_5(2x)(-3)^5 + C_6(-3)^6$

$(2 x - 3)^{6} = C_{0} (2 x)^{6} + C_{1} (2 x)^{5} (- 3) + C_{2} (2 x)^{4} (- 3)^{2} + C_{3} (2 x)^{3} (- 3)^{3} + C_{4} (2 x)^{2} (- 3)^{4} + C_{5} (2 x) (- 3)^{5} + C_{6} (- 3)^{6}$

Expanding this:

$= 64x^6 - 576x^5 + 2160x^4 - 4320x^3 + 4860x^2 - 2916x + 729$

$= 64 x^{6} - 576 x^{5} + 2160 x^{4} - 4320 x^{3} + 4860 x^{2} - 2916 x + 729$

5. Expand the expression

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

$(3 x + x 1)^{5}$

By using the Binomial Theorem, the expression

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

$(3 x + x 1)^{5}$ can be expanded as:

$\left(\frac{x}{3} + \frac{1}{x}\right)^5 = C_0 \left(\frac{x}{3}\right)^5 + C_1 \left(\frac{x}{3}\right)^4 \left(\frac{1}{x}\right) + C_2 \left(\frac{x}{3}\right)^3 \left(\frac{1}{x^2}\right) + \ldots$

$(3 x + x 1)^{5} = C_{0} (3 x)^{5} + C_{1} (3 x)^{4} (x 1) + C_{2} (3 x)^{3} (x ^{2} 1) + \dots$

Expanding this:

$= \frac{x^5}{243} + \frac{5x^3}{81} + \frac{10x}{27} + \frac{10}{9x} + \frac{5}{27x^3} + \frac{1}{243x^5}$

$= 243 x ^{5} + 815 x ^{3} + 2710 x + 9 x 10 + 27 x ^{3} 5 + 243 x ^{5} 1$

6. Using Binomial Theorem, evaluate

$(96)^3$

$(96)^{3}$

Using the Binomial Theorem:

$(96)^3 = (100 - 4)^3$

$(96)^{3} = (100 - 4)^{3}$

Expanding this:

$= 100^3 - 3(100)^2(4) + 3(100)(4)^2 - 4^3$

$= 10 0^{3} - 3 (100)^{2} (4) + 3 (100) (4)^{2} - 4^{3}$

Simplifying:

$= 1000000 - 120000 + 4800 - 64 = 884736$

7. Using Binomial Theorem, evaluate

$(102)^5$

$(102)^{5}$

Using the Binomial Theorem:

$(102)^5 = (100 + 2)^5$

$(102)^{5} = (100 + 2)^{5}$

Expanding this:

$= C_0(100)^5 + C_1(100)^4(2) + C_2(100)^3(2)^2 + \ldots$

$= C_{0} (100)^{5} + C_{1} (100)^{4} (2) + C_{2} (100)^{3} (2)^{2} + \dots$

Simplifying:

$= 10000000000 + 1000000000 + 40000000 + 80000 + 8000 + 32 = 11040808032$

8. Using Binomial Theorem, evaluate

$(101)^4$

$(101)^{4}$

Using the Binomial Theorem:

$(101)^4 = (100 + 1)^4$

$(101)^{4} = (100 + 1)^{4}$

Expanding this:

$= C_0(100)^4 + C_1(100)^3(1) + C_2(100)^2(1)^2 + \ldots$

$= C_{0} (100)^{4} + C_{1} (100)^{3} (1) + C_{2} (100)^{2} (1)^{2} + \dots$

Simplifying:

$= 100000000 + 4000000 + 60000 + 400 + 1 = 104060401$

9. Using Binomial Theorem, evaluate

$(99)^5$

$(99)^{5}$

Using the Binomial Theorem:

$(99)^5 = (100 - 1)^5$

$(99)^{5} = (100 - 1)^{5}$

Expanding this:

$= C_0(100)^5 - 5C_1(100)^4 + 10C_2(100)^3 - 10C_3(100)^2 + 5C_4(100) - C_5$

$= C_{0} (100)^{5} - 5 C_{1} (100)^{4} + 10 C_{2} (100)^{3} - 10 C_{3} (100)^{2} + 5 C_{4} (100) - C_{5}$

Simplifying:

$= 10000000000 - 500000000 + 10000000 - 100000 + 500 = 9509900499$

10. Using Binomial Theorem, indicate which number is larger

$(1.1)^{10000}$

$(1.1)^{10000}$ or 1000.

Expanding

$(1.1)^{10000}$

$(1.1)^{10000}$ using the Binomial Theorem:

$(1.1)^{10000} = 1 + 10000(0.1) + \ldots = 11000 + \text{other positive terms}$

$(1.1)^{10000} = 1 + 10000 (0.1) + \dots = 11000 + other positive terms$

Since it exceeds 1000, we conclude:

$(1.1)^{10000} > 1000$

11. Find

$(a + b)^4 - (a - b)^4$

$(a + b)^{4} - (a - b)^{4}$ . Hence, evaluate

$\sqrt{3} + \sqrt{2} - (\sqrt{3} - \sqrt{2})$

$3 + 2 - (3 - 2)$ .

By using Binomial Theorem, expand both

$(a + b)^4$

$(a + b)^{4}$ and

$(a - b)^4$

$(a - b)^{4}$ .

Expanding:

$(a + b)^4 = C_0a^4 + C_1a^3b + C_2a^2b^2 + C_3ab^3 + C_4b^4$

$(a + b)^{4} = C_{0} a^{4} + C_{1} a^{3} b + C_{2} a^{2} b^{2} + C_{3} a b^{3} + C_{4} b^{4}$

$(a - b)^4 = C_0a^4 - C_1a^3b + C_2a^2b^2 - C_3ab^3 + C_4b^4$

$(a - b)^{4} = C_{0} a^{4} - C_{1} a^{3} b + C_{2} a^{2} b^{2} - C_{3} a b^{3} + C_{4} b^{4}$

Now, subtract:

$(a + b)^4 - (a - b)^4 = 2C_1a^3b + 2C_3ab^3$

$(a + b)^{4} - (a - b)^{4} = 2 C_{1} a^{3} b + 2 C_{3} a b^{3}$

By substituting

$a = 3$

$a = 3$ and

$b = 2$

$b = 2$ , we get:

$= 8 \cdot 6 + 6 \cdot 2 = 48 + 12 = 60$

For

$\sqrt{3} + \sqrt{2} - (\sqrt{3} - \sqrt{2})$

$3 + 2 - (3 - 2)$ :

$= 8\sqrt{6}(3 + 2)$

$= 86 (3 + 2)$

$= 40\sqrt{6}$

$= 406$

12. Find

$(x + 1)^6 + (x - 1)^6$

$(x + 1)^{6} + (x - 1)^{6}$ . Hence, or otherwise evaluate

$\sqrt{2} + 1 + \sqrt{2} - 1$

$2 + 1 + 2 - 1$ .

By using the Binomial Theorem, expand both

$(x + 1)^6$

$(x + 1)^{6}$ and

$(x - 1)^6$

$(x - 1)^{6}$ .

Expanding:

$(x + 1)^6 = C_0x^6 + C_1x^5 + C_2x^4 + C_3x^3 + C_4x^2 + C_5x + C_6$

$(x + 1)^{6} = C_{0} x^{6} + C_{1} x^{5} + C_{2} x^{4} + C_{3} x^{3} + C_{4} x^{2} + C_{5} x + C_{6}$

$(x - 1)^6 = C_0x^6 - C_1x^5 + C_2x^4 - C_3x^3 + C_4x^2 - C_5x + C_6$

$(x - 1)^{6} = C_{0} x^{6} - C_{1} x^{5} + C_{2} x^{4} - C_{3} x^{3} + C_{4} x^{2} - C_{5} x + C_{6}$

Now, adding:

$(x + 1)^6 + (x - 1)^6 = 2(x^6 + 15x^4 + 20x^2 + 6)$

$(x + 1)^{6} + (x - 1)^{6} = 2 (x^{6} + 15 x^{4} + 20 x^{2} + 6)$

Substituting

$x = 2$

$x = 2$ :

$= 2(2^6 + 15 \cdot 2^4 + 20 \cdot 2^2 + 6)$

$= 2 (2^{6} + 15 \cdot 2^{4} + 20 \cdot 2^{2} + 6)$

$= 2(64 + 240 + 80 + 6) = 2 \times 390 = 780$

$= 2 (64 + 240 + 80 + 6) = 2 \times 390 = 780$

FAQs – Class 11 Maths Chapter 7 Exercise 7.1 Binomial Theorem

Q1. What is the focus of Exercise 7.1?
Exercise 7.1 focuses on understanding the Binomial Theorem for expanding binomial expressions and calculating individual terms in the expansion.

Q2. What is the Binomial Theorem?
The Binomial Theorem states that the expansion of

$(a + b)^n$

$(a + b)^{n}$ is given by the sum:

$(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r$

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

Where

$\binom{n}{r}$

$(r n)$ represents the binomial coefficient, also known as "n choose r."

Q3. What are the binomial coefficients?
The binomial coefficients

$\binom{n}{r}$

$(r n)$ are calculated using the formula:

$\binom{n}{r} = \frac{n!}{r!(n - r)!}$

$(r n) = r ! ( n - r )! n !$

These coefficients represent the number of ways to choose r elements from n.

Q4. Why is Exercise 7.1 important for exams?
Exercise 7.1 is crucial as it lays the foundation for understanding binomial expansions, which are frequently asked in Class 11 exams. It introduces key concepts that are vital for further topics like combinations and probability.

Q5. How can students prepare effectively for Exercise 7.1?
Students should:

Understand the general form of binomial expansion and the use of binomial coefficients.
Practice expanding different binomial expressions.
Work on identifying specific terms in the expansion using the binomial theorem.

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

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

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

FAQs – Class 11 Maths Chapter 7 Exercise 7.1 Binomial Theorem

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