How do you find the general term in a binomial expansion?

The (r+1)th term in (a + b)ⁿ is T(r+1) = nCr × aⁿ⁻ʳ × bʳ, where r ranges from 0 to n. Use this to find specific terms, the middle term, or the term with a particular power of x.

How do you find the middle term of a binomial expansion?

If n is even, there is one middle term: the (n/2 + 1)th term. If n is odd, there are two middle terms: the ((n+1)/2)th and ((n+3)/2)th terms.

Binomial Theorem — NCERT Class 11 Maths

Exercise Solutions Available

Step-by-step answers for all textbook questions in this chapter

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How to Use This Page

Attempt each NCERT exercise question independently before checking the solutions below. Every question from Chapter 7 (Binomial Theorem) is solved with full steps — expand using the theorem, identify the general term, and find the required term.

Key Concepts (Quick Recap)

Binomial Theorem: (a + b)ⁿ = Σ(nCr)(aⁿ⁻ʳ bʳ), r = 0 to n
General (r+1)th term: T(r+1) = nCr × aⁿ⁻ʳ × bʳ
Number of terms: n + 1
Middle term: If n even — one middle term T(n/2 + 1); if n odd — two middle terms T((n+1)/2) and T((n+3)/2)
Pascal's Triangle: nCr = (n−1)C(r−1) + (n−1)Cr
Term independent of x: Find r such that the power of x in T(r+1) equals 0

Exercise 7.1

Q1. Expand (u + v)⁵ using the binomial theorem.

Solution

(u + v)⁵ = ⁵C₀ u⁵ + ⁵C₁ u⁴v + ⁵C₂ u³v² + ⁵C₃ u²v³ + ⁵C₄ uv⁴ + ⁵C₅ v⁵

= u⁵ + 5u⁴v + 10u³v² + 10u²v³ + 5uv⁴ + v⁵

(Coefficients: ⁵C₀ = 1, ⁵C₁ = 5, ⁵C₂ = 10, ⁵C₃ = 10, ⁵C₄ = 5, ⁵C₅ = 1)

Q2. Expand (1 + x)⁵ using the binomial theorem.

Solution

(1 + x)⁵ = ⁵C₀ + ⁵C₁ x + ⁵C₂ x² + ⁵C₃ x³ + ⁵C₄ x⁴ + ⁵C₅ x⁵

= 1 + 5x + 10x² + 10x³ + 5x⁴ + x⁵

Q3. Expand (2 + x)⁵ using the binomial theorem.

Solution

(2 + x)⁵ = Σ⁵Cr (2)⁵⁻ʳ xʳ

= ⁵C₀(32) + ⁵C₁(16)x + ⁵C₂(8)x² + ⁵C₃(4)x³ + ⁵C₄(2)x⁴ + ⁵C₅ x⁵

= 32 + 80x + 80x² + 40x³ + 10x⁴ + x⁵

Q4. Expand (1 − 2x)⁵ using the binomial theorem.

Solution

(1 − 2x)⁵ = Σ⁵Cr (1)⁵⁻ʳ (−2x)ʳ

= 1 + 5(−2x) + 10(4x²) + 10(−8x³) + 5(16x⁴) + (−32x⁵)

= 1 − 10x + 40x² − 80x³ + 80x⁴ − 32x⁵

Q5. Expand (2x − 3)⁶ using the binomial theorem.

Solution

(2x − 3)⁶ = Σ⁶Cr (2x)⁶⁻ʳ (−3)ʳ

= ⁶C₀(64x⁶) + ⁶C₁(32x⁵)(−3) + ⁶C₂(16x⁴)(9) + ⁶C₃(8x³)(−27) + ⁶C₄(4x²)(81) + ⁶C₅(2x)(−243) + ⁶C₆(729)

= 64x⁶ − 6(32)(3)x⁵ + 15(16)(9)x⁴ − 20(8)(27)x³ + 15(4)(81)x² − 6(2)(243)x + 729

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

Q6. Expand (x² + 3/x)⁴ using the binomial theorem.

Solution

(x² + 3/x)⁴ = Σ⁴Cr (x²)⁴⁻ʳ (3/x)ʳ

= ⁴C₀ x⁸ + ⁴C₁ x⁶(3/x) + ⁴C₂ x⁴(9/x²) + ⁴C₃ x²(27/x³) + ⁴C₄ (81/x⁴)

= x⁸ + 4(3x⁵) + 6(9x²) + 4(27/x) + 81/x⁴

= x⁸ + 12x⁵ + 54x² + 108/x + 81/x⁴

Q7. Using binomial theorem, evaluate: (99)⁵.

Solution

99 = 100 − 1 (99)⁵ = (100 − 1)⁵

= ⁵C₀(100)⁵ − ⁵C₁(100)⁴ + ⁵C₂(100)³ − ⁵C₃(100)² + ⁵C₄(100) − ⁵C₅

= 10000000000 − 5(100000000) + 10(1000000) − 10(10000) + 5(100) − 1

= 10000000000 − 500000000 + 10000000 − 100000 + 500 − 1

= 9509900499

Q8. Using binomial theorem, evaluate: (102)⁶.

Solution

102 = 100 + 2 (102)⁶ = (100 + 2)⁶

= ⁶C₀(100)⁶ + ⁶C₁(100)⁵(2) + ⁶C₂(100)⁴(4) + ⁶C₃(100)³(8) + ⁶C₄(100)²(16) + ⁶C₅(100)(32) + ⁶C₆(64)

= 10¹² + 6(2)(10¹⁰) + 15(4)(10⁸) + 20(8)(10⁶) + 15(16)(10⁴) + 6(32)(100) + 64

= 10¹² + 12 × 10¹⁰ + 60 × 10⁸ + 160 × 10⁶ + 240 × 10⁴ + 19200 + 64

= 1000000000000 + 120000000000 + 6000000000 + 160000000 + 2400000 + 19200 + 64

= 1126162401264

Exercise 7.2

Q1. Find the coefficient of x⁵ in the expansion of (x + 1/x)⁸.

Solution

General term: T(r+1) = ⁸Cr x⁸⁻ʳ (1/x)ʳ = ⁸Cr x⁸⁻²ʳ

For x⁵: 8 − 2r = 5 ⟹ r = 3/2 — not an integer.

Wait, let me recheck. 8 − 2r = 5 ⟹ 2r = 3 ⟹ r = 1.5. Since r must be an integer, x⁵ does not appear.

Hmm, this can't be right for a standard NCERT problem. Let me re-examine.

If the problem is (x + 1/x)⁸, the powers of x are 8−2r for r = 0,1,...,8, giving even powers: 8, 6, 4, 2, 0, −2, −4, −6, −8. So x⁵ does not appear. Coefficient = 0.

The standard NCERT problem likely reads: Find the coefficient of x⁵ in (x + 3)⁸ or (1 + x)⁸ etc.

Alternative — Find coefficient of x⁵ in (1 + x)⁸: T(r+1) = ⁸Cr xʳ. For x⁵, r = 5. Coefficient = ⁸C5 = ⁸C3 = (8 × 7 × 6)/(3 × 2 × 1) = 56

Q2. Find the coefficient of x⁵ in the expansion of (1 + x)¹¹.

Solution

T(r+1) = ¹¹Cr xʳ. For x⁵: r = 5. Coefficient = ¹¹C5 = (11 × 10 × 9 × 8 × 7)/(5 × 4 × 3 × 2 × 1) = 55440/120 = 462

Q3. Find the coefficient of a⁴b² in the expansion of (a + 2b)⁶.

Solution

T(r+1) = ⁶Cr a⁶⁻ʳ (2b)ʳ = ⁶Cr (2ʳ) a⁶⁻ʳ bʳ

For a⁴b²: 6 − r = 4 ⟹ r = 2 Coefficient = ⁶C₂ × 2² = 15 × 4 = 60

Q4. Find the coefficient of a⁵ in the expansion of (2a − 1)⁹.

Solution

T(r+1) = ⁹Cr (2a)⁹⁻ʳ (−1)ʳ = ⁹Cr (2)⁹⁻ʳ (−1)ʳ a⁹⁻ʳ

For a⁵: 9 − r = 5 ⟹ r = 4 Coefficient = ⁹C₄ × 2⁵ × (−1)⁴ = 126 × 32 × 1 = 4032

Q5. Find the 4th term in the expansion of (x − 2y)¹².

Solution

T₄ = T(3+1) = ¹²C₃ x⁹ (−2y)³ = (12 × 11 × 10)/(3 × 2 × 1) × x⁹ × (−8y³) = 220 × (−8) x⁹y³ = −1760x⁹y³

Q6. Find the 7th term in the expansion of (3x − 1/2x)¹¹.

Solution

T₇ = T(6+1) = ¹¹C₆ (3x)⁵ (−1/2x)⁶ = 462 × 243x⁵ × 1/(64x⁶) = 462 × 243/(64) × 1/x = 112266/64 × 1/x = 56133/(32x) or equivalently −56133/(32x)

Wait — (−1/2x)⁶ = (+1)/(64x⁶) since 6 is even. So the sign is positive.

T₇ = ⁶C₆ × (3)⁵ × (1/2)⁶ × x⁵/x⁶ Wait, let me redo carefully.

T(r+1) = ¹¹Cr (3x)¹¹⁻ʳ (−1/(2x))ʳ

For T₇, r = 6: = ¹¹C₆ (3x)⁵ (−1/(2x))⁶ = 462 × 243x⁵ × 1/(64x⁶) = 462 × 243/(64x) = 112266/(64x) = 56133/(32x)

Q7. Find the middle term in the expansion of (2x − 1/x)⁹.

Solution

n = 9 (odd), so there are two middle terms: T₅ and T₆.

T₅ (r = 4): ⁹C₄ (2x)⁵ (−1/x)⁴ = 126 × 32x⁵ × 1/x⁴ = 126 × 32x = 4032x

T₆ (r = 5): ⁹C₅ (2x)⁴ (−1/x)⁵ = 126 × 16x⁴ × (−1/x⁵) = 126 × 16 × (−1/x) = −2016/x

Q8. Find the middle term in the expansion of (1 + x)⁸.

Solution

n = 8 (even), so one middle term: T₅ (the (8/2 + 1) = 5th term).

T₅ (r = 4) = ⁸C₄ x⁴ = 70x⁴

Middle term = 70x⁴

Q9. Find the middle term in the expansion of (x/3 + 9y)¹⁰.

Solution

n = 10 (even), so one middle term: T₆ (the (10/2 + 1) = 6th term).

T₆ (r = 5) = ¹⁰C₅ (x/3)⁵ (9y)⁵ = 252 × (x⁵/243) × (59049y⁵) = 252 × 59049/243 × x⁵y⁵ = 252 × 243 × x⁵y⁵ Wait: 59049/243 = 243 (since 243² = 59049). So: = 252 × 243 × x⁵y⁵ = 61236x⁵y⁵

Q10. Find the term independent of x in the expansion of (x + 1/x)⁸.

Solution

T(r+1) = ⁸Cr x⁸⁻ʳ (1/x)ʳ = ⁸Cr x⁸⁻²ʳ

For term independent of x: 8 − 2r = 0 ⟹ r = 4

T₅ = ⁸C₄ = 70

Miscellaneous Exercise

Q1. Find the 13th term in the expansion of (9x − 1/3√x)¹⁸.

Solution

T₁₃ = T(12+1) = ¹⁸C₁₂ (9x)⁶ (−1/(3√x))¹²

Wait, let me be more careful. (9x)^(18−12) = (9x)⁶ and (−1/(3√x))¹²

= ¹⁸C₁₂ (9x)⁶ (−1/(3x^(1/2)))¹² = ¹⁸C₁₂ × 9⁶ × x⁶ × (−1)¹²/(3¹² × x⁶) = ¹⁸C₁₂ × 9⁶/3¹² × x⁶/x⁶ = ¹⁸C₁₂ × (9/3²)⁶ × 1

Note: 9 = 3², so 9⁶ = 3¹², giving 9⁶/3¹² = 1.

= ¹⁸C₁₂ × 1 = ¹⁸C₁₂ = ¹⁸C₆ = (18 × 17 × 16 × 15 × 14 × 13)/(6 × 5 × 4 × 3 × 2 × 1) = (18 × 17 × 16 × 15 × 14 × 13)/720

Numerator: 18 × 17 = 306; 306 × 16 = 4896; 4896 × 15 = 73440; 73440 × 14 = 1028160; 1028160 × 13 = 13366080 13366080/720 = 18564

Q2. Find the coefficient of x⁵ in the expansion of (1 + x²)⁵ (1 + x)⁴.

Solution

(1 + x²)⁵ = 1 + 5x² + 10x⁴ + 10x⁶ + 5x⁸ + x¹⁰

(1 + x)⁴ = 1 + 4x + 6x² + 4x³ + x⁴

For x⁵ from the product:

From (1 + x²)⁵: 1, and from (1 + x)⁴: x⁵ — but max degree is x⁴. No contribution.
From 5x² × (4x³) = 20x⁵
From 10x⁴ × x⁴ ... wait, (1+x)⁴ gives x⁴ at most. So:
1 × (coefficient of x⁵ in (1+x)⁴) = 0 (since max degree is 4)
5x² × 4x³ = 20x⁵
10x⁴ × x⁴ ... no, need x⁵ not x⁸.

Wait — 10x⁴ from first times (coefficient of x¹ from second): but (1+x)⁴ has no x¹ term? Actually it does: 4x. So 10x⁴ × 4x = 40x⁵.

Total coefficient of x⁵ = 0 + 20 + 40 = 60

Q3. If the coefficients of x² and x³ in the expansion of (1 + x)ⁿ are in the ratio 1:2, find n.

Solution

Coefficient of x² = nC2 = n(n−1)/2 Coefficient of x³ = nC3 = n(n−1)(n−2)/6

Ratio: [n(n−1)/2] / [n(n−1)(n−2)/6] = 1/2

(6)/(2(n−2)) = 1/2 3/(n−2) = 1/2 6 = n − 2 n = 8

Q4. If the coefficient of xᵏ in (1 + x)ⁿ is 36, the coefficient of xᵏ⁺¹ is 84 and the coefficient of xᵏ⁺² is 126, find n.

Solution

nCk = 36, nC(k+1) = 84, nC(k+2) = 126.

Using nC(k+1)/nCk = (n−k)/(k+1): 84/36 = (n−k)/(k+1) 7/3 = (n−k)/(k+1) ... (1)

Using nC(k+2)/nC(k+1) = (n−k−1)/(k+2): 126/84 = (n−k−1)/(k+2) 3/2 = (n−k−1)/(k+2) ... (2)

From (1): 7(k+1) = 3(n−k) ⟹ 7k + 7 = 3n − 3k ⟹ 10k + 7 = 3n ... (3) From (2): 3(k+2) = 2(n−k−1) ⟹ 3k + 6 = 2n − 2k − 2 ⟹ 5k + 8 = 2n ⟹ n = (5k+8)/2 ... (4)

Substituting (4) into (3): 10k + 7 = 3(5k+8)/2 20k + 14 = 15k + 24 5k = 10 ⟹ k = 2

n = (5(2) + 8)/2 = 18/2 = n = 9

Q5. Find a positive value of m for which the coefficient of x² in the expansion of (1 + x)²ᵐ is equal to the coefficient of x³ in (1 + x)²ᵐ⁺¹.

Solution

Coefficient of x² in (1 + x)²ᵐ = (2m)C2 = 2m(2m−1)/2 Coefficient of x³ in (1 + x)²ᵐ⁺¹ = (2m+1)C3 = (2m+1)(2m)(2m−1)/6

Setting equal: 2m(2m−1)/2 = (2m+1)(2m)(2m−1)/6

Dividing both sides by 2m(2m−1) (assuming m > 0): 1/2 = (2m+1)/6 3 = 2m + 1 m = 1

Q6. The coefficients of the (r−1)th, rth and (r+1)th terms in the expansion of (1 + x)ⁿ are in the ratio 1:3:5. Find n and r.

Solution

Coefficient of (r−1)th term = nC(r−2) Coefficient of rth term = nC(r−1) Coefficient of (r+1)th term = nCr

nC(r−2) : nC(r−1) : nCr = 1 : 3 : 5

From first ratio: nC(r−1)/nC(r−2) = 3 (n−r+2)/(r−1) = 3 ⟹ n − r + 2 = 3r − 3 ⟹ n + 5 = 4r ... (1)

From second ratio: nCr/nC(r−1) = 5/3 (n−r+1)/r = 5/3 ⟹ 3n − 3r + 3 = 5r ⟹ 3n + 3 = 8r ... (2)

From (1): r = (n+5)/4. Substituting into (2): 3n + 3 = 8(n+5)/4 = 2(n+5) 3n + 3 = 2n + 10 n = 7

r = (7+5)/4 = 3. r = 3

Q7. Prove that the coefficient of the middle term in the expansion of (1 + x)²ⁿ is equal to the sum of the coefficients of two middle terms in (1 + x)²ⁿ⁻¹.

Solution

For (1 + x)²ⁿ, n is even. Middle term = T(n+1) with coefficient = (2n)Cn.

For (1 + x)²ⁿ⁻¹, n−1 is odd (since 2n−1 is odd). Two middle terms: Tₙ and T(n+1). Coefficients = (2n−1)C(n−1) and (2n−1)Cn.

Sum = (2n−1)C(n−1) + (2n−1)Cn = (2n)Cn (by Pascal's rule: (n−1)C(r−1) + (n−1)Cr = nCr)

This equals the coefficient of the middle term of (1 + x)²ⁿ. ✓

Q8. Find the term independent of x in the expansion of (x + 1/x)¹².

Solution

T(r+1) = ¹²Cr x¹²⁻ʳ (1/x)ʳ = ¹²Cr x¹²⁻²ʳ

For independent of x: 12 − 2r = 0 ⟹ r = 6

T₇ = ¹²C₆ = (12 × 11 × 10 × 9 × 8 × 7)/(6 × 5 × 4 × 3 × 2 × 1) = 924

Term independent of x = 924

Internal Links

[Chapter 7 Guide — Binomial Theorem](/school/cbse/class/11/maths/cbse-binomial-theorem)
[All Class 11 Maths Solutions](/school/cbse/class/11/maths/solutions)
[Class 11 Maths Subject Guide](/school/cbse/class/11/maths)

Frequently Asked Questions

The binomial theorem states that (a + b)ⁿ = Σ(nCr)(aⁿ⁻ʳ)(bʳ) for r = 0 to n, where n is a positive integer. Each term is (nCr)(aⁿ⁻ʳ bʳ) and there are n+1 terms in the expansion.