NCERT Solutions for Class 11 Maths Chapter 7: Binomial Theorem – Free PDF Download

Chapter 7 of the current (2026-27 rationalised) NCERT Class 11 Maths textbook, Binomial Theorem, teaches students how to expand expressions of the form (a + b)ⁿ for any positive integer n without multiplying the binomial out term by term. The chapter builds the theorem from Pascal’s Triangle, states it formally with binomial coefficients (ⁿC_r), and applies it to expand algebraic expressions, approximate large powers of numbers close to a round figure, and prove divisibility results. These Class 11 Maths Chapter 7 solutions are also useful as quick revision notes before exams.

Exercise 7.1

Q1. Expand the expression (1 – 2x)⁵

By the Binomial Theorem,

(1 – 2x)⁵ = ⁵C₀(1)⁵ – ⁵C₁(1)⁴(2x) + ⁵C₂(1)³(2x)² – ⁵C₃(1)²(2x)³ + ⁵C₄(1)(2x)⁴ – ⁵C₅(2x)⁵
= 1 – 5(2x) + 10(4x²) – 10(8x³) + 5(16x⁴) – 32x⁵
= 1 – 10x + 40x² – 80x³ + 80x⁴ – 32x⁵

Q2. Expand the expression (2/x – x/2)⁵

By the Binomial Theorem, the general term is ⁵C_k(2/x)^5–k(–x/2)^k. Evaluating for k = 0 to 5:

(2/x – x/2)⁵ = 32/x⁵ – 40/x³ + 20/x – 5x + (5/8)x³ – (1/32)x⁵

Q3. Expand the expression (2x – 3)⁶

(2x – 3)⁶ = ⁶C₀(2x)⁶ – ⁶C₁(2x)⁵(3) + ⁶C₂(2x)⁴(3)² – ⁶C₃(2x)³(3)³ + ⁶C₄(2x)²(3)⁴ – ⁶C₅(2x)(3)⁵ + ⁶C₆(3)⁶
= 64x⁶ – 576x⁵ + 2160x⁴ – 4320x³ + 4860x² – 2916x + 729

Q4. Expand the expression (x/3 + 1/x)⁵

(x/3 + 1/x)⁵ = x⁵/243 + 5x³/81 + 10x/27 + 10/(9x) + 5/(3x³) + 1/x⁵

Q5. Expand the expression (x + 1/x)⁶

(x + 1/x)⁶ = ⁶C₀x⁶ + ⁶C₁x⁵(1/x) + ⁶C₂x⁴(1/x)² + ⁶C₃x³(1/x)³ + ⁶C₄x²(1/x)⁴ + ⁶C₅x(1/x)⁵ + ⁶C₆(1/x)⁶
= x⁶ + 6x⁴ + 15x² + 20 + 15/x² + 6/x⁴ + 1/x⁶

Using binomial theorem, evaluate each of the following:

Q6. (96)³

96 = 100 – 4, so (96)³ = (100 – 4)³
= ³C₀(100)³ – ³C₁(100)²(4) + ³C₂(100)(4)² – ³C₃(4)³
= 1000000 – 120000 + 4800 – 64
= 884736

Q7. (102)⁵

102 = 100 + 2, so (102)⁵ = (100 + 2)⁵
= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32
= 11040808032

Q8. (101)⁴

101 = 100 + 1, so (101)⁴ = (100 + 1)⁴
= 100000000 + 4000000 + 60000 + 400 + 1
= 104060401

Q9. (99)⁵

99 = 100 – 1, so (99)⁵ = (100 – 1)⁵
= 10000000000 – 500000000 + 10000000 – 100000 + 500 – 1
= 9509900499

Q10. Using Binomial Theorem, indicate which number is larger (1.1)¹⁰⁰⁰⁰ or 1000.

(1.1)¹⁰⁰⁰⁰ = (1 + 0.1)¹⁰⁰⁰⁰ = ¹⁰⁰⁰⁰C₀ + ¹⁰⁰⁰⁰C₁(0.1) + other positive terms
= 1 + 10000 × 0.1 + other positive terms = 1 + 1000 + other positive terms > 1001
Since all the remaining terms in the expansion are positive, (1.1)¹⁰⁰⁰⁰ > 1001 > 1000.
Hence, (1.1)¹⁰⁰⁰⁰ is larger than 1000.

Q11. Find (a + b)⁴ – (a – b)⁴. Hence, evaluate (√3 + √2)⁴ – (√3 – √2)⁴.

(a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴
(a – b)⁴ = a⁴ – 4a³b + 6a²b² – 4ab³ + b⁴
Subtracting, the even-power terms cancel:
(a + b)⁴ – (a – b)⁴ = 8a³b + 8ab³
Putting a = √3, b = √2:
= 8(√3)³(√2) + 8(√3)(√2)³ = 8(3√3)(√2) + 8(√3)(2√2) = 24√6 + 16√6 = 40√6

Q12. Find (x + 1)⁶ + (x – 1)⁶. Hence or otherwise evaluate (√2 + 1)⁶ + (√2 – 1)⁶.

(x + 1)⁶ = x⁶ + 6x⁵ + 15x⁴ + 20x³ + 15x² + 6x + 1
(x – 1)⁶ = x⁶ – 6x⁵ + 15x⁴ – 20x³ + 15x² – 6x + 1
Adding, the odd-power terms cancel:
(x + 1)⁶ + (x – 1)⁶ = 2x⁶ + 30x⁴ + 30x² + 2
Putting x = √2 (so x² = 2, x⁴ = 4, x⁶ = 8):
= 2(8) + 30(4) + 30(2) + 2 = 16 + 120 + 60 + 2 = 198

Q13. Show that 9ⁿ⁺¹ – 8n – 9 is divisible by 64, whenever n is a positive integer.

Write 9ⁿ⁺¹ = (1 + 8)ⁿ⁺¹. By the Binomial Theorem,
9ⁿ⁺¹ = ⁿ⁺¹C₀ + ⁿ⁺¹C₁(8) + ⁿ⁺¹C₂(8)² + ⁿ⁺¹C₃(8)³ + … + ⁿ⁺¹C_n+1(8)ⁿ⁺¹
= 1 + 8(n + 1) + 8²[ⁿ⁺¹C₂ + ⁿ⁺¹C₃(8) + … + ⁿ⁺¹C_n+1(8)^n–1]
= 1 + 8n + 8 + 64k, where k = ⁿ⁺¹C₂ + ⁿ⁺¹C₃(8) + … + ⁿ⁺¹C_n+1(8)^n–1 is a natural number
Therefore, 9ⁿ⁺¹ – 8n – 9 = 64k, which is divisible by 64 for every positive integer n.

Q14. Prove that Σ_r=0ⁿ 3^{r n}C_r = 4ⁿ

By the Binomial Theorem, (a + b)ⁿ = Σ_r=0^{n n}C_r a^n–rb^r. Putting a = 1, b = 3:
(1 + 3)ⁿ = Σ_r=0^{n n}C_r(1)^n–r(3)^r = Σ_r=0ⁿ 3^{r n}C_r
Since 1 + 3 = 4, this gives Σ_r=0ⁿ 3^{r n}C_r = 4ⁿ, as required.

Miscellaneous Exercise

Q1. If a and b are distinct integers, prove that a – b is a factor of aⁿ – bⁿ, whenever n is a positive integer. [Hint: write aⁿ = (a – b + b)ⁿ and expand]

Write a = (a – b) + b, so aⁿ = [(a – b) + b]ⁿ. By the Binomial Theorem,
aⁿ = ⁿC₀(a – b)ⁿ + ⁿC₁(a – b)^n–1b + … + ⁿC_n–1(a – b)b^n–1 + ⁿC_nbⁿ
aⁿ = (a – b)ⁿ + ⁿC₁(a – b)^n–1b + … + ⁿC_n–1(a – b)b^n–1 + bⁿ
So, aⁿ – bⁿ = (a – b)[(a – b)^n–1 + ⁿC₁(a – b)^n–2b + … + ⁿC_n–1b^n–1]
The expression in brackets is an integer, so (a – b) divides aⁿ – bⁿ for every positive integer n. Hence proved.

Q2. Evaluate (√3 + √2)⁶ – (√3 – √2)⁶.

Only the odd-power terms of b survive in (a + b)⁶ – (a – b)⁶:
(a + b)⁶ – (a – b)⁶ = 2[⁶C₁a⁵b + ⁶C₃a³b³ + ⁶C₅ab⁵]
With a = √3, b = √2: a⁵ = 9√3, a³ = 3√3, b³ = 2√2, b⁵ = 4√2
= 2[6(9√3)(√2) + 20(3√3)(2√2) + 6(√3)(4√2)]
= 2[54√6 + 120√6 + 24√6] = 2(198√6) = 396√6

Q3. Find the value of (a² + √(a² – 1))⁴ + (a² – √(a² – 1))⁴.

Let x = a² and y = √(a² – 1), so y² = a² – 1. Only the even-power terms of y survive:
(x + y)⁴ + (x – y)⁴ = 2[⁴C₀x⁴ + ⁴C₂x²y² + ⁴C₄y⁴] = 2[x⁴ + 6x²y² + y⁴]
= 2[a⁸ + 6a⁴(a² – 1) + (a² – 1)²]
= 2[a⁸ + 6a⁶ – 6a⁴ + a⁴ – 2a² + 1]
= 2a⁸ + 12a⁶ – 10a⁴ – 4a² + 2

Q4. Find an approximation of (0.99)⁵ using the first three terms of its expansion.

(0.99)⁵ = (1 – 0.01)⁵
≈ ⁵C₀(1)⁵ – ⁵C₁(1)⁴(0.01) + ⁵C₂(1)³(0.01)²
= 1 – 5(0.01) + 10(0.0001)
= 1 – 0.05 + 0.001 = 0.951

Q5. Expand using Binomial Theorem (1 + x/2 – 2/x)⁴, x ≠ 0.

Grouping as [(1 + x/2) + (–2/x)]⁴ and expanding fully using the Binomial Theorem (each power of (1 + x/2) itself expanded again), and collecting like powers of x:

(1 + x/2 – 2/x)⁴ = x⁴/16 + x³/2 + x²/2 – 4x – 5 + 16/x + 8/x² – 32/x³ + 16/x⁴

Q6. Find the expansion of (3x² – 2ax + 3a²)³ using binomial theorem.

Treating 3x², –2ax and 3a² as the three terms of a trinomial and expanding (A + B + C)³ = A³ + B³ + C³ + 3A²B + 3A²C + 3AB² + 3B²C + 3AC² + 3BC² + 6ABC, and collecting terms by power of x and a:

(3x² – 2ax + 3a²)³ = 27x⁶ – 54ax⁵ + 117a²x⁴ – 116a³x³ + 117a⁴x² – 54a⁵x + 27a⁶

Class 11 Maths Chapter 7 – Notes and Extra Questions

• Binomial Theorem (positive integral index): (a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁a^n–1b + ⁿC₂a^n–2b² + … + ⁿC_nbⁿ = Σ_k=0^{n n}C_ka^n–kb^k.
• General term: T_r+1 = ⁿC_ra^n–rb^r — this is used to find any particular term without expanding the whole binomial.
• Number of terms: the expansion of (a + b)ⁿ always has exactly (n + 1) terms.
• Pascal’s Triangle: the array of binomial coefficients ⁿC_r, where each entry is the sum of the two entries above it; also known as Meru Prastara (given by Pingala).
• Special expansions: (x – y)ⁿ = ⁿC₀xⁿ – ⁿC₁x^n–1y + … + (–1)^{n n}C_nyⁿ; (1 + x)ⁿ = ⁿC₀ + ⁿC₁x + … + ⁿC_nxⁿ; putting x = 1 gives ⁿC₀ + ⁿC₁ + … + ⁿC_n = 2ⁿ.
• Index/power symmetry: in every term of the expansion, the sum of the powers of a and b always equals n.
• Common technique: (a + b)ⁿ ± (a – b)ⁿ is used to isolate the odd-power or even-power terms only — very useful for irrational/surd evaluation questions.
• Approximation technique: for numbers close to a round base (like 98, 101, 0.99), write the number as (round base ± small number) and use only the first few terms of the expansion for a quick approximate value.

Frequently Asked Questions

How many exercises are there in the current (2026-27) NCERT Class 11 Maths Chapter 7?

In the rationalised NCERT textbook currently prescribed (Reprint 2026-27), Chapter 7 (Binomial Theorem) has only one exercise, Exercise 7.1 (14 questions), followed by the Miscellaneous Exercise (6 questions). Earlier, pre-rationalisation editions of this chapter had an additional Exercise 7.2 and a section on the general/rational-index binomial theorem, both of which have been removed from the current syllabus.

What is the Binomial Theorem formula for Class 11?

For any positive integer n, (a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁a^n–1b + ⁿC₂a^n–2b² + … + ⁿC_nbⁿ. This can also be written compactly using sigma notation as Σ_k=0^{n n}C_ka^n–kb^k.

How do you quickly find a large power like 102⁵ or 99⁵ using the Binomial Theorem?

Write the number as a sum or difference of a round number and a small number — for example 102 = 100 + 2, or 99 = 100 – 1 — then expand using the Binomial Theorem. Since most of the resulting terms involve powers of 100, they are easy to compute and add or subtract, avoiding repeated multiplication.

Is the Binomial Theorem for any rational index still in the current NCERT Class 11 Maths syllabus?

No. In the current (2026-27) rationalised edition, Chapter 7 covers the Binomial Theorem only for positive integral indices. The topic of the binomial theorem for any rational index, which appeared in older editions of this chapter, has been removed from the present syllabus.