NCERT Solutions for Class 11 Maths Chapter 7: Binomial Theorem (NCERT 2026–27)

These Class 11 Maths Chapter 7 solutions cover the complete Binomial Theorem chapter from the NCERT textbook (Reprint 2026–27). Every question of Exercise 7.1 and the Miscellaneous Exercise on Chapter 7 is reproduced verbatim and solved step by step — binomial expansions, evaluation of powers like (96)3 and (102)5, comparisons, divisibility proofs and the sum of binomial coefficients — all verified against the book’s answer key.

Class: 11 Subject: Mathematics Chapter: 7 – Binomial Theorem Exercises: Exercise 7.1, Miscellaneous Exercise Session: 2026–27
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Chapter 7 Overview

Chapter 7 of Class 11 Maths, Binomial Theorem, gives a quick way to expand (a + b)n for any positive integer n without repeated multiplication. It begins with Pascal’s triangle (the Meru-Prastara of Pingala), rewrites its entries using combinations nCr, and proves the binomial theorem for positive integral indices by mathematical induction. The chapter then explores special cases such as (x − y)n, (1 + x)n and (1 − x)n, the meaning of binomial coefficients, and applications including evaluating large powers, comparing numbers, and proving divisibility results. The Class 11 Maths Chapter 7 solutions below work through every textbook question with full reasoning.

Key Concepts & Definitions

Binomial: an algebraic expression with two terms, such as a + b or 2x − 3.

Binomial coefficient: the number nCr = n! / [r!(n − r)!], 0 ≤ r ≤ n, that multiplies each term in the expansion of (a + b)n.

Pascal’s triangle: a triangular array in which each entry is the sum of the two entries above it; the n-th row gives the coefficients of (a + b)n.

General term: the (r + 1)-th term in the expansion of (a + b)n is Tr+1 = nCr an−r br.

Number of terms: the expansion of (a + b)n has exactly (n + 1) terms — one more than the index.

Sum of indices: in every term of (a + b)n the indices of a and b add up to n.

Important Formulas (Chapter 7)

Binomial theorem: (a + b)n = nC0an + nC1an−1b + nC2an−2b2 + … + nCnbn.

Difference form: (x − y)n = nC0xn − nC1xn−1y + nC2xn−2y2 − … + (−1)n nCnyn.

Special cases: (1 + x)n = nC0 + nC1x + … + nCnxn;   (1 − x)n = nC0 − nC1x + … + (−1)n nCnxn.

General (r + 1)-th term: Tr+1 = nCr an−r br.

Sum of coefficients: putting x = 1,  nC0 + nC1 + … + nCn = 2n;   with alternating signs the sum is 0.

Pascal’s rule: nCr−1 + nCr = n+1Cr.

Exercise 7.1 Solutions

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified against the answers given at the back of the book.

Expand each of the expressions in Exercises 1 to 5.

1. (1 − 2x)5

SOLUTION Using (a + b)5 with a = 1, b = −2x: (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 − 2x)5 = 1 − 10x + 40x2 − 80x3 + 80x4 − 32x5.

2. (2/x − x/2)5

SOLUTION Take a = 2/x, b = −x/2 and expand by the binomial theorem (power 5): = 5C0(2/x)5 + 5C1(2/x)4(−x/2) + 5C2(2/x)3(−x/2)2 + 5C3(2/x)2(−x/2)3 + 5C4(2/x)(−x/2)4 + 5C5(−x/2)5. Term 1: 32/x5.   Term 2: 5 · (16/x4) · (−x/2) = −40/x3.   Term 3: 10 · (8/x3) · (x2/4) = 20/x. Term 4: 10 · (4/x2) · (−x3/8) = −5x.   Term 5: 5 · (2/x) · (x4/16) = 5x3/8.   Term 6: −x5/32. ∴ (2/x − x/2)5 = 32/x5 − 40/x3 + 20/x − 5x + (5/8)x3 − (1/32)x5.

3. (2x − 3)6

SOLUTION Take a = 2x, b = −3, n = 6. Coefficients 6Cr = 1, 6, 15, 20, 15, 6, 1. (2x)6 = 64x6; 6(2x)5(−3) = 6(32x5)(−3) = −576x5; 15(2x)4(9) = 15(16x4)(9) = 2160x4. 20(2x)3(−27) = 20(8x3)(−27) = −4320x3; 15(2x)2(81) = 15(4x2)(81) = 4860x2; 6(2x)(−243) = −2916x; (−3)6 = 729. ∴ (2x − 3)6 = 64x6 − 576x5 + 2160x4 − 4320x3 + 4860x2 − 2916x + 729.

4. (x/3 + 1/x)5

SOLUTION Take a = x/3, b = 1/x, n = 5; coefficients 1, 5, 10, 10, 5, 1. (x/3)5 = x5/243.   5(x/3)4(1/x) = 5(x4/81)(1/x) = 5x3/81.   10(x/3)3(1/x)2 = 10(x3/27)(1/x2) = 10x/27. 10(x/3)2(1/x)3 = 10(x2/9)(1/x3) = 10/(9x).   5(x/3)(1/x)4 = 5/(3x3).   (1/x)5 = 1/x5. ∴ (x/3 + 1/x)5 = x5/243 + (5/81)x3 + (10/27)x + 10/(9x) + 5/(3x3) + 1/x5.

5. (x + 1/x)6

SOLUTION Take a = x, b = 1/x, n = 6; coefficients 1, 6, 15, 20, 15, 6, 1. x6 + 6x5(1/x) + 15x4(1/x2) + 20x3(1/x3) + 15x2(1/x4) + 6x(1/x5) + (1/x6). ∴ (x + 1/x)6 = x6 + 6x4 + 15x2 + 20 + 15/x2 + 6/x4 + 1/x6.

Using binomial theorem, evaluate each of the following:

6. (96)3

SOLUTION Write 96 = 100 − 4. Then (96)3 = (100 − 4)3 = 1003 − 3(100)2(4) + 3(100)(4)2 − 43. = 1000000 − 120000 + 4800 − 64 = 884736.

7. (102)5

SOLUTION Write 102 = 100 + 2. Then (102)5 = (100 + 2)5. = 1005 + 5(100)4(2) + 10(100)3(2)2 + 10(100)2(2)3 + 5(100)(2)4 + 25. = 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32. ∴ (102)5 = 11040808032.

8. (101)4

SOLUTION Write 101 = 100 + 1. Then (101)4 = (100 + 1)4. = 1004 + 4(100)3(1) + 6(100)2(1)2 + 4(100)(1)3 + 14. = 100000000 + 4000000 + 60000 + 400 + 1 = 104060401.

9. (99)5

SOLUTION Write 99 = 100 − 1. Then (99)5 = (100 − 1)5. = 1005 − 5(100)4(1) + 10(100)3(1)2 − 10(100)2(1)3 + 5(100)(1)4 − 15. = 10000000000 − 500000000 + 10000000 − 100000 + 500 − 1 = 9509900499.

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

SOLUTION Write 1.1 = 1 + 0.1, so (1.1)10000 = (1 + 0.1)10000. By the binomial theorem this equals 10000C0 + 10000C1(0.1) + (other positive terms) = 1 + 10000(0.1) + (positive terms). = 1 + 1000 + (positive terms) > 1001 > 1000. ∴ (1.1)10000 > 1000.

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

SOLUTION (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 = 8ab(a2 + b2) = 8(a3b + ab3). Put a = √3, b = √2: a2 = 3, b2 = 2, ab = √6. = 8(√6)(3 + 2) = 8 × 5 × √6 = 40√6.

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

SOLUTION (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 = 2(x6 + 15x4 + 15x2 + 1). Put x = √2: x2 = 2, x4 = 4, x6 = 8. = 2(8 + 15×4 + 15×2 + 1) = 2(8 + 60 + 30 + 1) = 2(99) = 198.

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

SOLUTION Write 9 = 1 + 8, so 9n+1 = (1 + 8)n+1. By the binomial theorem: (1 + 8)n+1 = n+1C0 + n+1C1(8) + n+1C2(8)2 + n+1C3(8)3 + … = 1 + 8(n + 1) + 82[n+1C2 + n+1C3(8) + …] = 1 + 8n + 8 + 64k, where k = n+1C2 + n+1C3(8) + … is a natural number. ∴ 9n+1 − 8n − 9 = (9 + 8n + 64k) − 8n − 9 = 64k, which is divisible by 64 for every positive integer n.

14. Prove that ∑r=0n 3r nCr = 4n.

SOLUTION Recall the special case (1 + x)n = nC0 + nC1x + nC2x2 + … + nCnxn = ∑r=0n nCr xr. Put x = 3: ∑r=0n nCr 3r = (1 + 3)n = 4n. ∴ ∑r=0n 3r nCr = 4n. ✓

Miscellaneous Exercise on Chapter 7 — Solutions

1. 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]

SOLUTION Write a = (a − b) + b. Then an = [(a − b) + b]n. Expand by the binomial theorem: an = nC0(a − b)n + nC1(a − b)n−1b + … + nCn−1(a − b)bn−1 + nCnbn. The last term is bn. Bringing it across: an − bn = nC0(a − b)n + nC1(a − b)n−1b + … + nCn−1(a − b)bn−1. Every remaining term contains (a − b) as a factor, so an − bn = (a − b) · k, where k = [nC0(a − b)n−1 + … + nCn−1bn−1] is an integer. ∴ (a − b) is a factor of an − bn for every positive integer n. ✓

2. Evaluate (√3 + √2)6 − (√3 − √2)6.

SOLUTION Let a = √3, b = √2. Using (a + b)6 − (a − b)6, the even-power terms cancel and the odd ones double: (a + b)6 − (a − b)6 = 2[6a5b + 20a3b3 + 6ab5] = 2ab[6a4 + 20a2b2 + 6b4]. With a2 = 3, b2 = 2: a4 = 9, b4 = 4, a2b2 = 6, ab = √6. = 2√6 [6(9) + 20(6) + 6(4)] = 2√6 [54 + 120 + 24] = 2√6 (198) = 396√6.

3. Find the value of (a2 + √(a2 − 1))4 + (a2 − √(a2 − 1))4.

SOLUTION Let x = a2 and y = √(a2 − 1), so y2 = a2 − 1. We need (x + y)4 + (x − y)4. Adding the two expansions, odd-power terms cancel: (x + y)4 + (x − y)4 = 2[x4 + 6x2y2 + y4]. Now x2 = a4, x4 = a8, y2 = a2 − 1, y4 = (a2 − 1)2 = a4 − 2a2 + 1. x4 + 6x2y2 + y4 = a8 + 6a4(a2 − 1) + (a4 − 2a2 + 1) = a8 + 6a6 − 6a4 + a4 − 2a2 + 1 = a8 + 6a6 − 5a4 − 2a2 + 1. Multiply by 2: ∴ the value is 2a8 + 12a6 − 10a4 − 4a2 + 2.

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

SOLUTION Write 0.99 = 1 − 0.01, so (0.99)5 = (1 − 0.01)5. First three terms: 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 = 0.9510 (approx.).

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

SOLUTION Group as [(1 + x/2) − 2/x]4. Let A = 1 + x/2, B = 2/x and use (A − B)4 = A4 − 4A3B + 6A2B2 − 4AB3 + B4. A4 = (1 + x/2)4 = 1 + 4(x/2) + 6(x/2)2 + 4(x/2)3 + (x/2)4 = 1 + 2x + (3/2)x2 + (1/2)x3 + x4/16. −4A3B: A3 = (1 + x/2)3 = 1 + (3/2)x + (3/4)x2 + x3/8. Then −4(2/x)A3 = −(8/x)[1 + (3/2)x + (3/4)x2 + x3/8] = −8/x − 12 − 6x − x2. 6A2B2: A2 = 1 + x + x2/4, B2 = 4/x2. Then 6(4/x2)[1 + x + x2/4] = (24/x2)[1 + x + x2/4] = 24/x2 + 24/x + 6. −4AB3: B3 = 8/x3. Then −4(1 + x/2)(8/x3) = −(32/x3)(1 + x/2) = −32/x3 − 16/x2. B4 = (2/x)4 = 16/x4. Add all five parts and collect like terms: Constants: 1 − 12 + 6 = −5.   x: 2x − 6x = −4x.   x2: (3/2)x2 − x2 = (1/2)x2.   x3: (1/2)x3.   x4: x4/16. 1/x: −8/x + 24/x = 16/x.   1/x2: 24/x2 − 16/x2 = 8/x2.   1/x3: −32/x3.   1/x4: 16/x4. ∴ (1 + x/2 − 2/x)4 = 16/x4 − 32/x3 + 8/x2 + 16/x − 4x + (1/2)x2 + (1/2)x3 + x4/16 − 5.

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

SOLUTION Group as [(3x2 − 2ax) + 3a2]3. Let P = 3x2 − 2ax, Q = 3a2; use (P + Q)3 = P3 + 3P2Q + 3PQ2 + Q3. P3 = (3x2 − 2ax)3: = (3x2)3 − 3(3x2)2(2ax) + 3(3x2)(2ax)2 − (2ax)3 = 27x6 − 54ax5 + 36a2x4 − 8a3x3. 3P2Q: P2 = (3x2 − 2ax)2 = 9x4 − 12ax3 + 4a2x2. Then 3(3a2)P2 = 9a2(9x4 − 12ax3 + 4a2x2) = 81a2x4 − 108a3x3 + 36a4x2. 3PQ2: Q2 = 9a4. Then 3(3x2 − 2ax)(9a4) = 27a4(3x2 − 2ax) = 81a4x2 − 54a5x. Q3 = (3a2)3 = 27a6. Now add and collect like terms by power of x: x6: 27x6.   x5: −54ax5.   x4: 36a2x4 + 81a2x4 = 117a2x4.   x3: −8a3x3 − 108a3x3 = −116a3x3. x2: 36a4x2 + 81a4x2 = 117a4x2.   x: −54a5x.   constant: 27a6. ∴ (3x2 − 2ax + 3a2)3 = 27x6 − 54ax5 + 117a2x4 − 116a3x3 + 117a4x2 − 54a5x + 27a6.

Common Mistakes to Avoid

Watch out for these

  • Forgetting the alternating signs in (a − b)n — every odd-position term carries a minus sign.
  • Not raising the whole bracket to its power: e.g. (−2x)4 = 16x4, not −16x4, and (2/x)3 = 8/x3.
  • Using the wrong row of coefficients — the expansion of (a + b)n has n + 1 terms, so power 6 needs 7 coefficients.
  • In divisibility proofs, choosing the split badly — write the base as 1 + (multiple) so most terms collect into the required factor.
  • Dropping the “other positive terms” argument in inequality problems — you only need enough terms to beat the bound.
  • Arithmetic slips when collecting like terms in trinomial expansions; group by power of x and check each coefficient.

Practice MCQs & Assertion–Reason

1. The number of terms in the expansion of (a + b)n is:

(a) n    (b) n − 1    (c) n + 1    (d) 2n

2. The general (r + 1)-th term in the expansion of (a + b)n is:

(a) nCr ar bn−r    (b) nCr an−r br    (c) nCr−1 an−r br    (d) nCr an br

3. The value of nC0 + nC1 + nC2 + … + nCn is:

(a) 0    (b) n    (c) 2n    (d) n2

4. Using the binomial theorem, (101)4 equals:

(a) 104060401    (b) 104040601    (c) 101010101    (d) 104060001

5. The coefficient of x5 in the expansion of (2x − 3)6 is:

(a) 2160    (b) −576    (c) −4320    (d) 64

6. The value of ∑r=0n nCr 2r is:

(a) 2n    (b) 3n    (c) 4n    (d) 1

7. The value of nCr−1 + nCr is:

(a) nCr+1    (b) n+1Cr    (c) n−1Cr    (d) n+1Cr+1

8. The middle term in the expansion of (x + 1/x)6 is:

(a) 15    (b) 20    (c) 6    (d) 1

9. The term independent of x in (x + 1/x)6 is:

(a) 6    (b) 15    (c) 20    (d) 1

10. Comparing the two numbers, (1.1)10000 and 1000:

(a) (1.1)10000 < 1000    (b) (1.1)10000 = 1000    (c) (1.1)10000 > 1000    (d) cannot be decided

Answer key: 1-(c), 2-(b), 3-(c), 4-(a), 5-(b), 6-(b), 7-(b), 8-(b), 9-(c), 10-(c).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The expansion of (2x − 3)6 has 7 terms.

Reason: The number of terms in (a + b)n is one more than the index n.

A-R 2. Assertion: nC0 + nC1 + … + nCn = 2n.

Reason: Putting x = 1 in (1 + x)n = ∑ nCrxr gives the sum of all binomial coefficients.

A-R 3. Assertion: (a + b)4 − (a − b)4 = 8ab(a2 + b2).

Reason: When (a − b)4 is subtracted from (a + b)4, the terms with even powers of b cancel.

A-R 4. Assertion: 9n+1 − 8n − 9 is divisible by 64 for every positive integer n.

Reason: Writing 9 = 1 + 8 and expanding (1 + 8)n+1 leaves a remainder term that is a multiple of 82 = 64.

A-R 5. Assertion: A perfect square can be evaluated using the binomial theorem, e.g. (96)3 = 884736.

Reason: The general term of (a + b)n is Tr+1 = nCr ar bn−r.

Answer key: 1-(A), 2-(A), 3-(A), 4-(A), 5-(C).

Quick Revision Summary

  • (a + b)n = ∑r=0n nCr an−r br, valid for every positive integer n.
  • The expansion has (n + 1) terms; in each term the indices of a and b add up to n.
  • The (r + 1)-th term is Tr+1 = nCr an−r br.
  • Special cases: (1 + x)n, (1 − x)n and (x − y)n; the coefficients nCr form Pascal’s triangle.
  • Sum of all coefficients = 2n; alternating sum = 0; Pascal’s rule: nCr−1 + nCr = n+1Cr.
  • Use a clever split (such as 100 ± small, or 1 + multiple) to evaluate big powers, compare numbers, or prove divisibility.

How to score full marks in this chapter

For evaluation questions, always express the base as a round number ± a small number (e.g. 96 = 100 − 4) so each binomial term is easy to compute. For “hence evaluate” problems, first derive the algebraic identity, then substitute the surds — this earns method marks even if arithmetic slips. In divisibility and inequality proofs, state the binomial expansion clearly and identify the term that gives the required factor or beats the bound. Show every coefficient when expanding, and collect like terms power by power so no term is lost.

Frequently Asked Questions

What is Class 11 Maths Chapter 7 Binomial Theorem about?

Chapter 7 explains how to expand (a + b)n for any positive integer n using binomial coefficients nCr and Pascal’s triangle. It covers the binomial theorem and its proof by induction, special cases such as (1 + x)n, and applications like evaluating large powers, comparing numbers and proving divisibility.

How many exercises are there in Class 11 Maths Chapter 7?

There are two exercise sets — Exercise 7.1 (14 questions) and the Miscellaneous Exercise on Chapter 7 (6 questions). Every question of both is solved step by step on this page.

What is the general term in the binomial expansion?

The general or (r + 1)-th term of (a + b)n is Tr+1 = nCr an−r br. Choosing the right value of r lets you find a particular term, the term independent of x, or the middle term.

Are these Class 11 Maths Chapter 7 solutions free?

Yes. All solutions are free and follow the official NCERT Mathematics textbook for the 2026–27 session, with every answer verified against the book’s answer key.

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