CBSE Class 11 Maths Chapter 8: Binomial Theorem NCERT Solutions

Reviewed by NCERT Help Editorial Team · Updated on 29 Jul 2026

NCERT Solutions PDF Class 11 PDF

CBSE Class 11 Mathematics chapter 8, the Binomial Theorem, introduces the expansion of expressions like (x+y)^n and (x-y)^n. NCERT Solutions for this chapter break down the binomial theorem formula and the properties of binomial coefficients. Students will learn how to calculate individual terms in an expansion, simplify powers, and combine terms. The solutions emphasize understanding the binomial expansion formula, calculating combinations (nCr), and correctly applying alternating signs when dealing with subtractions. This chapter is vital for mastering algebraic manipulations and is a key topic for exam success, building a solid foundation for advanced mathematics.

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Quick info

Board	CBSE
Class	Class 11
Subject	गणित
Session	2026
Language	English
Type	NCERT Solutions
Chapter	8. द्विपद प्रमेय

Chapter summary

This chapter focuses on the Binomial Theorem, specifically addressing exercises on expanding binomial expressions. It details the application of the binomial expansion formula for positive integer exponents. Students will learn to systematically calculate each term, utilizing combinations and powers of the binomial terms. The solutions cover expansions for both positive and negative terms within the binomial, emphasizing the correct application of signs and coefficients.

Learning outcomes

Understand the binomial theorem for expanding expressions of the form (x+y)^n.
Apply the binomial theorem to expand expressions with negative terms, like (x-y)^n.
Calculate binomial coefficients (nCr) accurately.
Systematically expand binomial expressions up to a given power.
Simplify and present the final expanded form of binomial expressions.

Topics covered

Paper topics

Binomial Theorem
Expansion of Binomial Expressions
Binomial Coefficients
Combinations (nCr)
General Term in Binomial Expansion
Expansion of (x+y)^n
Expansion of (x-y)^n
Algebraic Expressions
Polynomial Expansion

Important topics

Binomial Theorem Formula
Calculating nCr
Expansion of (x+y)^n
Expansion of (x-y)^n
Simplifying Expanded Terms

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Questions and Solutions

Question 1

Expand the expression: (1 - 2x)⁵

Solution:

We use the binomial theorem for the expansion of (a+b)^n, which is given by:

$(a+b)^n = {}^nC_0 a^n + {}^nC_1 a^{n-1}b + {}^nC_2 a^{n-2}b^2 + \dots + {}^nC_n b^n$

In this case, a = 1, b = -2x, and n = 5.

Applying the formula:

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

Now, we calculate the binomial coefficients and simplify each term:

${}^5C_0 = 1$
${}^5C_1 = 5$
${}^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$
${}^5C_3 = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$
${}^5C_4 = {}^5C_1 = 5$
${}^5C_5 = {}^5C_0 = 1$

Substitute these values back into the expansion:

$(1 - 2x)^5 = 1 \times 1 + 5 \times 1 \times (-2x) + 10 \times 1 \times (4x^2) + 10 \times 1 \times (-8x^3) + 5 \times 1 \times (16x^4) + 1 \times (-32x^5)$

Simplify further:

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

The expanded form of (1 - 2x)⁵ is $1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5$ .

Question 2

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

Solution:

We use the binomial theorem for the expansion of (a+b)^n, where a = $\frac{2}{y}$ , b = $-\frac{x}{2}$ , and n = 5.

The expansion is:

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

We know the binomial coefficients: ${}^5C_0=1, {}^5C_1=5, {}^5C_2=10, {}^5C_3=10, {}^5C_4=5, {}^5C_5=1$ .

Now, let's calculate each term:

Term 1: $1 \times \left(\frac{32}{y^5}\right) = \frac{32}{y^5}$
Term 2: $5 \times \left(\frac{16}{y^4}\right) \times \left(-\frac{x}{2}\right) = -\frac{40x}{y^4}$
Term 3: $10 \times \left(\frac{8}{y^3}\right) \times \left(\frac{x^2}{4}\right) = \frac{20x^2}{y^3}$
Term 4: $10 \times \left(\frac{4}{y^2}\right) \times \left(-\frac{x^3}{8}\right) = -\frac{5x^3}{2y^2}$
Term 5: $5 \times \left(\frac{2}{y}\right) \times \left(\frac{x^4}{16}\right) = \frac{5x^4}{8y}$
Term 6: $1 \times \left(-\frac{x^5}{32}\right) = -\frac{x^5}{32}$

Combining these terms, the expansion is:

$(\frac{2}{y} - \frac{x}{2})^{5} = \frac{32}{y^{5}} - \frac{40 x}{y^{4}} + \frac{20 x^{2}}{y^{3}} - \frac{5 x^{3}}{2 y^{2}} + \frac{5 x^{4}}{8 y} - \frac{x^{5}}{32}$

Question 3

Expand the expression: (2x - 3)⁶

Solution:

We use the binomial theorem to expand (a+b)^n, with a = 2x, b = -3, and n = 6.

The expansion is:

$(2x - 3)^6 = {}^6C_0 (2x)^6 + {}^6C_1 (2x)^5 (-3) + {}^6C_2 (2x)^4 (-3)^2 + {}^6C_3 (2x)^3 (-3)^3 + {}^6C_4 (2x)^2 (-3)^4 + {}^6C_5 (2x)^1 (-3)^5 + {}^6C_6 (-3)^6$

First, calculate the binomial coefficients:

${}^6C_0 = 1$
${}^6C_1 = 6$
${}^6C_2 = \frac{6 \times 5}{2 \times 1} = 15$
${}^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$
${}^6C_4 = {}^6C_2 = 15$
${}^6C_5 = {}^6C_1 = 6$
${}^6C_6 = {}^6C_0 = 1$

Now, calculate the powers of the terms:

$(2x)^6 = 64x^6$
$(2x)^5 = 32x^5$
$(2x)^4 = 16x^4$
$(2x)^3 = 8x^3$
$(2x)^2 = 4x^2$
$(2x)^1 = 2x$
$(-3)^2 = 9$
$(-3)^3 = -27$
$(-3)^4 = 81$
$(-3)^5 = -243$
$(-3)^6 = 729$

Substitute these values into the expansion:

$(2x - 3)^6 = 1(64x^6) + 6(32x^5)(-3) + 15(16x^4)(9) + 20(8x^3)(-27) + 15(4x^2)(81) + 6(2x)(-243) + 1(729)$

Simplify each term:

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

The expanded form of (2x - 3)⁶ is $64x^6 - 576x^5 + 2160x^4 - 4320x^3 + 4860x^2 - 2916x + 729$ .

Question 4

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

Solution:

We use the binomial theorem to expand $(a+b)^n$ , with $a = \frac{x}{3}$ , $b = \frac{1}{x}$ , and $n = 5$ .

The expansion is:

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

The binomial coefficients are: ${}^5C_0=1, {}^5C_1=5, {}^5C_2=10, {}^5C_3=10, {}^5C_4=5, {}^5C_5=1$ .

Now, we calculate each term:

Term 1: $1 \times \frac{x^5}{243} = \frac{x^5}{243}$
Term 2: $5 \times \frac{x^4}{81} \times \frac{1}{x} = \frac{5x^3}{81}$
Term 3: $10 \times \frac{x^3}{27} \times \frac{1}{x^2} = \frac{10x}{27}$
Term 4: $10 \times \frac{x^2}{9} \times \frac{1}{x^3} = \frac{10}{9x}$
Term 5: $5 \times \frac{x}{3} \times \frac{1}{x^4} = \frac{5}{3x^3}$
Term 6: $1 \times \frac{1}{x^5} = \frac{1}{x^5}$

Combining these terms, the expansion is:

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

Question 5

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

Solution:

We use the binomial theorem to expand $(a+b)^n$ , with $a = x$ , $b = \frac{1}{x}$ , and $n = 6$ .

The expansion is:

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

The binomial coefficients are:

${}^6C_0 = 1$
${}^6C_1 = 6$
${}^6C_2 = 15$
${}^6C_3 = 20$
${}^6C_4 = 15$
${}^6C_5 = 6$
${}^6C_6 = 1$

Now, simplify each term by combining the powers of x:

Term 1: $1 \times x^6 = x^6$
Term 2: $6 \times x^5 \times \frac{1}{x} = 6x^4$
Term 3: $15 \times x^4 \times \frac{1}{x^2} = 15x^2$
Term 4: $20 \times x^3 \times \frac{1}{x^3} = 20$
Term 5: $15 \times x^2 \times \frac{1}{x^4} = \frac{15}{x^2}$
Term 6: $6 \times x \times \frac{1}{x^5} = \frac{6}{x^4}$
Term 7: $1 \times \frac{1}{x^6} = \frac{1}{x^6}$

Combining these terms, the expansion is:

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

Common mistakes

Errors in calculating binomial coefficients (nCr).
Incorrectly applying the alternating signs in expansions of (x-y)^n.
Mistakes in simplifying powers of terms, especially with fractions or negative exponents.
Forgetting to expand powers of the constants or coefficients within the binomial terms.

Revision tips

Review the binomial expansion formula and the properties of nCr before starting.
Practice calculating nCr values for different n and r.
Pay close attention to the signs when expanding expressions with a minus sign.
Work through each step of the expansion carefully to avoid calculation errors.
Verify your final answer by checking the pattern of powers and coefficients.

Practice MCQs

Q1. What is the general formula for the binomial expansion of (x+y)^n?

Q2. In the expansion of (a-b)^n, what is the sign of the term involving $b^3$ ?

Q3. Which of the following is the value of 5C2?

Q4. What is the first term in the expansion of (2x - 3)^6?

Q5. The expansion of (x/3 + 1/x)^3 involves terms with powers of x. What is the highest power of x in the expansion?

Frequently asked questions

What is the Binomial Theorem?

The Binomial Theorem provides a formula for expanding expressions of the form (x+y)^n, where n is a non-negative integer. It states that (x+y)^n = Σ (nCr) * x^(n-r) * y^r, for r from 0 to n.

How do I expand (x-y)^n using the Binomial Theorem?

To expand (x-y)^n, you can treat it as (x + (-y))^n. The terms will alternate in sign because of the negative sign in front of y. The formula becomes (x-y)^n = Σ (nCr) * x^(n-r) * (-y)^r, for r from 0 to n.

What are binomial coefficients?

Binomial coefficients, denoted as nCr or C(n,r), are the numerical coefficients in the expansion of a binomial power. They represent the number of ways to choose r items from a set of n items and are calculated using the formula nCr = n! / (r! * (n-r)!).

How can these NCERT solutions help with exam preparation?

These solutions provide clear, step-by-step explanations for each problem, helping you understand the application of the Binomial Theorem. Practicing with these examples reinforces the concepts and calculation methods needed for exams.

What is the purpose of expanding binomial expressions?

Expanding binomial expressions simplifies them into a polynomial form, which can be easier to analyze, differentiate, integrate, or use in further algebraic manipulations.

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CBSE Class 11 Maths Chapter 8: Binomial Theorem NCERT Solutions

Quick info

Chapter summary

Learning outcomes

Topics covered

Paper topics

Important topics

PDF preview

Questions and Solutions

Question 1

Question 2

Question 3

Question 4

Question 5

Common mistakes

Revision tips

Practice MCQs

Frequently asked questions

Important links