Home › NCERT Solutions › Class 11 Maths › Chapter 7

NCERT Solutions · Class 11 Maths · Chapter 7

Chapter 7: Binomial Theorem (Algebra)

Step-by-step solutions to all exercises of Chapter 7, Binomial Theorem (NCERT Class 11 Maths, 2026-27): binomial expansions, evaluating powers like (96)³ and (99)⁵, divisibility proofs and the Miscellaneous Exercise. All 20 questions are answered, with the key answer highlighted.

NCERT textbook Chapter 7 (PDF)

Exam revision? Chapter 7 notes with key points, diagrams and practice questions: see the free sample · ₹199 for all 14 chapters

Free NCERT solutions by Notes Bazar · www.notesbazar.in/ncert-solutions/class-11-maths/chapter-7-binomial-theorem

$(a + b)^{n} = \sum_{r = 0}^{n}^{n} C_{r} a^{n - r} b^{r}$ ; general term $T_{r + 1} =^{n} C_{r} a^{n - r} b^{r}$ . Write 96 = 100 − 4 or 102 = 100 + 2 to evaluate powers.

Exercise 7.1

Expand (1 − 2x)⁵.

Solution

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

1 − 10x + 40x² − 80x³ + 80x⁴ − 32x⁵

Expand (2/x − x/2)⁵.

Solution

Terms ⁵C_(k)(2/x)⁵⁻ᵏ(−x/2)ᵏ: k = 0: 32/x⁵; k = 1: 5(16/x⁴)(−x/2) = −40/x³; k = 2: 10(8/x³)(x²/4) = 20/x; k = 3: 10(4/x²)(−x³/8) = −5x; k = 4: 5(2/x)(x⁴/16) = 5x³/8; k = 5: −x⁵/32. 32/x⁵ − 40/x³ + 20/x − 5x + 5x³/8 − x⁵/32.

32/x⁵ − 40/x³ + 20/x − 5x + 5x³/8 − x⁵/32

Expand (2x − 3)⁶.

Solution

(2x)⁶ − 6(2x)⁵3 + 15(2x)⁴9 − 20(2x)³27 + 15(2x)²81 − 6(2x)243 + 729 = 64x⁶ − 576x⁵ + 2160x⁴ − 4320x³ + 4860x² − 2916x + 729.

64x⁶ − 576x⁵ + 2160x⁴ − 4320x³ + 4860x² − 2916x + 729

Expand (x/3 + 1/x)⁵.

Solution

(x/3)⁵ + 5(x/3)⁴(1/x) + 10(x/3)³(1/x²) + 10(x/3)²(1/x³) + 5(x/3)(1/x⁴) + 1/x⁵ = x⁵/243 + 5x³/81 + 10x/27 + 10/(9x) + 5/(3x³) + 1/x⁵.

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

Expand (x + 1/x)⁶.

Solution

Coefficients 1, 6, 15, 20, 15, 6, 1 with powers x⁶, x⁴, …: x⁶ + 6x⁴ + 15x² + 20 + 15/x² + 6/x⁴ + 1/x⁶.

x⁶ + 6x⁴ + 15x² + 20 + 15/x² + 6/x⁴ + 1/x⁶

Evaluate: (96)³

Solution

(100 − 4)³ = 1000000 − 3(10000)(4) + 3(100)(16) − 64 = 1000000 − 120000 + 4800 − 64 = 884736.

884736

Evaluate: (102)⁵

Solution

(100 + 2)⁵ = 10¹⁰ + 5(10⁸)(2) + 10(10⁶)(4) + 10(10⁴)

See the working above.

Evaluate: (101)⁴

Solution

+ 5(100)(16) + 32 = 11040808032. (8) (100 + 1)⁴ = 10⁸ + 4(10⁶) + 6(10⁴) + 4(100) + 1 = 104060401.

11040808032; 104060401

Evaluate: (99)⁵.

Solution

(100 − 1)⁵ = 10¹⁰ − 5(10⁸) + 10(10⁶) − 10(10⁴) + 5(100) − 1 = 9509900499.

9509900499

Which is larger, (1.1)¹⁰⁰⁰⁰ or 1000?

Solution

(1 + 0.1)¹⁰⁰⁰⁰ = 1 + 10000 × 0.1 + (other positive terms) = 1001 + … > 1000: (1.1)¹⁰⁰⁰⁰ is larger.

(1.1)¹⁰⁰⁰⁰ is larger

Find (a + b)⁴ − (a − b)⁴. Hence evaluate (√3 + √2)⁴ − (√3 − √2)⁴.

Solution

The even-power terms of b cancel: = 2[4a³b + 4ab³] = 8ab(a² + b²). With a = √3, b = √2: 8√6 (3 + 2) = 40√6.

8ab(a² + b²); 40√6

Find (x + 1)⁶ + (x − 1)⁶. Hence evaluate (√2 + 1)⁶ + (√2 − 1)⁶.

Solution

The odd-power terms cancel: = 2[x⁶ + 15x⁴ + 15x² + 1]. At x = √2: 2[8 + 60 + 30 + 1] = 198.

198

Show that 9ⁿ⁺¹ − 8n − 9 is divisible by 64 for every positive integer n.

Solution

9ⁿ⁺¹ = (1 + 8)ⁿ⁺¹ = 1 + (n + 1)8 + ⁿ⁺¹C₂8² + ⁿ⁺¹C₃8³ + … + 8ⁿ⁺¹. So 9ⁿ⁺¹ − 8n − 9 = 8²[ⁿ⁺¹C₂ + ⁿ⁺¹C₃8 + … + 8ⁿ⁻¹] = 64k, k a natural number: divisible by 64.

64k

Prove that

\sum_{r = 0}^{n} 3^{r}^{n} C_{r} = 4^{n}

Solution

By the theorem, (1 + x)ⁿ = Σ ⁿCᵣxʳ. Put x = 3: Σ 3ʳⁿCᵣ = (1 + 3)ⁿ = 4ⁿ.

4ⁿ

Miscellaneous Exercise on Chapter 7

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

Solution

aⁿ = [(a − b) + b]ⁿ = ⁿC₀(a − b)ⁿ + ⁿC₁(a − b)ⁿ⁻¹b + … + ⁿCₙ₋₁(a − b)bⁿ⁻¹ + bⁿ. So aⁿ − bⁿ = (a − b)[(a − b)ⁿ⁻¹ + ⁿC₁(a − b)ⁿ⁻²b + … + ⁿCₙ₋₁bⁿ⁻¹] = (a − b)k, with k an integer: a − b divides aⁿ − bⁿ.

a − b divides aⁿ − bⁿ

Evaluate (√3 + √2)⁶ − (√3 − √2)⁶.

Solution

(a + b)⁶ − (a − b)⁶ = 2[6a⁵b + 20a³b³ + 6ab⁵]. With a = √3, b = √2: a⁵b = 9√6, a³b³ = 6√6, ab⁵ = 4√6. = 2[54 + 120 + 24]√6 = 396√6.

396√6

Find (a² + √(a² − 1))⁴ + (a² − √(a² − 1))⁴.

Solution

With A = a², B = √(a² − 1): (A + B)⁴ + (A − B)⁴ = 2[A⁴ + 6A²B² + B⁴] = 2[a⁸ + 6a⁴(a² − 1) + (a² − 1)²] = 2[a⁸ + 6a⁶ − 5a⁴ − 2a² + 1] = 2a⁸ + 12a⁶ − 10a⁴ − 4a² + 2.

2a⁸ + 12a⁶ − 10a⁴ − 4a² + 2

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

Solution

(1 − 0.01)⁵ ≈ 1 − 5(0.01) + 10(0.01)² = 1 − 0.05 + 0.001 = 0.951 (exact 0.95099…).

0.951

Expand (1 + x/2 − 2/x)⁴, x ≠ 0.

Solution

Let u = 1 + x/2, v = −2/x: (u + v)⁴ = u⁴ + 4u³v + 6u²v² + 4uv³ + v⁴. u⁴ = 1 + 2x + 3x²/2 + x³/2 + x⁴/16; 4u³v = −(8/x)(1 + 3x/2 + 3x²/4 + x³/8) = −8/x − 12 − 6x − x²; 6u²v² = (24/x²)(1 + x + x²/4) = 24/x² + 24/x + 6; 4uv³ = −(32/x³)(1 + x/2) = −32/x³ − 16/x²; v⁴ = 16/x⁴. Adding: x⁴/16 + x³/2 + x²/2 − 4x − 5 + 16/x + 8/x² − 32/x³ + 16/x⁴.

x⁴/16 + x³/2 + x²/2 − 4x − 5 + 16/x + 8/x² − 32/x³ + 16/x⁴

Find the expansion of (3x² − 2ax + 3a²)³.

Solution

Write it as [3(x² + a²) − 2ax]³ = 27(x² + a²)³ − 3 × 9(x² + a²)²(2ax) + 3 × 3(x² + a²)(4a²x²) − 8a³x³. Expanding each and collecting powers of x: 27x⁶ − 54ax⁵ + 117a²x⁴ − 116a³x³ + 117a⁴x² − 54a⁵x + 27a⁶.

27x⁶ − 54ax⁵ + 117a²x⁴ − 116a³x³ + 117a⁴x² − 54a⁵x + 27a⁶

Cover of Complete Maths Notes for Class 11 (NCERT 2026-27)

Done with the answers?Now revise the whole chapter: Class 11 Mathematics notes

For NCERT Mathematics Class 11 (2026-27 edition) - all 14 chapters
Typed concept notes with every formula, theorem and method, with worked examples
Every NCERT exercise and miscellaneous question solved step by step

₹199₹24920% off · all 14 chapters · instant PDF

Read free sample

What is inside the Chapter 7 notes →

← Chapter 6: Permutations and Combinations Chapter 8: Sequences and Series →

More free NCERT solutions for Class 11Class 11 Physics · 14 chapters Class 11 Chemistry · 9 chapters Class 11 Biology · 19 chapters All classes →

Found a mistake or need help with a question? Message us on WhatsApp. Free study tips and new solutions: join our WhatsApp Channel.