NCERT Class 11 · Mathematics
Chapter 7: Binomial Theorem
Master Pascal's triangle, binomial expansion for positive integers, general term, middle term and applications.
📋 25 MCQs
🔢 15 Numericals
⭐ Moderate to Hard
⏱ 90 Minutes
SECTION A
Multiple Choice Questions
1
The number of terms in the expansion of (a + b)ⁿ is:
(A) n
(B) n − 1
(C) n + 1
(D) 2n
✓ ANSWER: (C)
The expansion of (a+b)ⁿ has
(n+1) terms — one more than the index.
2
The general term (r+1)th term in the expansion of (a + b)ⁿ is:
(A) ⁿCᵣ aⁿ⁻ʳ bʳ
(B) ⁿCᵣ aʳ bⁿ⁻ʳ
(C) ⁿCᵣ aⁿ bʳ
(D) ⁿCᵣ₋₁ aⁿ⁻ʳ bʳ
✓ ANSWER: (A)
T(r+1) =
ⁿCᵣ · aⁿ⁻ʳ · bʳ
3
The binomial coefficients in (a + b)⁴ are:
(A) 1, 3, 3, 1
(B) 1, 4, 4, 1
(C) 1, 4, 6, 4, 1
(D) 1, 5, 10, 10, 5, 1
✓ ANSWER: (C)
⁴C₀, ⁴C₁, ⁴C₂, ⁴C₃, ⁴C₄ =
1, 4, 6, 4, 1
4
In the expansion of (a + b)ⁿ, the sum of powers of a and b in each term equals:
(A) r
(B) n + 1
(C) n
(D) 2n
✓ ANSWER: (C)
In each term T(r+1) = ⁿCᵣ aⁿ⁻ʳ bʳ, the sum of powers = (n−r) + r =
n.
5
The expansion of (1 + x)ⁿ at x = 1 gives:
(A) 1
(B) 0
(C) 2ⁿ
(D) n!
✓ ANSWER: (C)
(1+1)ⁿ = ⁿC₀ + ⁿC₁ + ⁿC₂ + ... + ⁿCₙ =
2ⁿ
6
The middle term in the expansion of (a + b)⁶ is the:
(A) 3rd term
(B) 4th term
(C) 5th term
(D) 6th term
✓ ANSWER: (B)
For even n, middle term = ((n/2)+1)th term = (6/2+1) =
4th term
7
Which term is the middle term in the expansion of (x + y)⁵?
(A) 2nd term
(B) 3rd term
(C) 4th term
(D) 5th term
✓ ANSWER: (B)
For odd n=5, middle term = ((n+1)/2)th = (5+1)/2 =
3rd term
8
The 3rd term in the expansion of (x + 2)⁶ is:
(A) 60x⁴
(B) 240x³
(C) 60x⁴
(D) 12x⁵
✓ ANSWER: (A)
T₃ = T(2+1) = ⁶C₂ · x⁴ · 2² = 15 × x⁴ × 4 =
60x⁴
9
In the expansion of (x − y)ⁿ, the signs of terms alternate. The rth term has sign:
(A) Always positive
(B) Always negative
(C) (−1)ʳ⁻¹
(D) (−1)ʳ
✓ ANSWER: (C)
T(r+1) = ⁿCᵣ xⁿ⁻ʳ (−y)ʳ = (−1)ʳ ⁿCᵣ xⁿ⁻ʳ yʳ
So the rth term (r starting from 1) has sign
(−1)ʳ⁻¹.
10
Pascal's triangle gets each interior number by:
(A) Multiplying the two numbers above it
(B) Adding the two numbers above it
(C) Subtracting the two numbers above it
(D) Dividing the two numbers above it
✓ ANSWER: (B)
Each entry = sum of the two entries directly above it. This is Pascal's identity: ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ.
11
The value of ⁿC₀ − ⁿC₁ + ⁿC₂ − ... + (−1)ⁿ ⁿCₙ is:
✓ ANSWER: (C)
Putting x = 1 in (1−x)ⁿ = ⁿC₀ − ⁿC₁ + ⁿC₂ − ... gives (1−1)ⁿ =
0
12
The coefficient of x³ in (1 + x)⁵ is:
(A) 5
(B) 10
(C) 15
(D) 20
✓ ANSWER: (B)
Coefficient of x³ = ⁵C₃ = 5!/(3!2!) =
10
13
The expansion of (a + b)⁴ is:
(A) a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴
(B) a⁴ + 3a³b + 3a²b² + ab³ + b⁴
(C) a⁴ + 4a³b + 4a²b² + 4ab³ + b⁴
(D) a⁴ + b⁴
✓ ANSWER: (A)
Using ⁴C₀=1, ⁴C₁=4, ⁴C₂=6, ⁴C₃=4, ⁴C₄=1:
a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴
14
The first three terms of (1.01)¹⁰⁰⁰⁰⁰⁰ using binomial theorem give approximately:
(A) 10001
(B) 10000
(C) Greater than 10000
(D) Less than 10000
✓ ANSWER: (C)
(1.01)¹⁰⁰⁰⁰⁰⁰ = (1+0.01)¹⁰⁰⁰⁰⁰⁰ ≥ 1 + 1000000×0.01 = 1 + 10000 = 10001 > 10000
So it is
greater than 10000.
15
The term independent of x in (x + 1/x)⁶ is:
(A) ⁶C₂
(B) ⁶C₃
(C) ⁶C₄
(D) 1
✓ ANSWER: (B)
T(r+1) = ⁶Cᵣ x⁶⁻ʳ (1/x)ʳ = ⁶Cᵣ x⁶⁻²ʳ
For term independent of x: 6−2r = 0 ⟹ r = 3
Term = ⁶C₃ =
20
16
The binomial theorem states (a + b)ⁿ where the index n must be:
(A) Any real number
(B) Any rational number
(C) Positive integer (in this chapter)
(D) Negative integer
✓ ANSWER: (C)
In this chapter, the Binomial Theorem is studied for
positive integral indices only.
17
The coefficient of x⁴ in (x + 2)⁶ is:
(A) 15
(B) 60
(C) 240
(D) 12
✓ ANSWER: (B)
T(r+1) = ⁶Cᵣ x⁶⁻ʳ 2ʳ. For x⁴: 6−r=4 ⟹ r=2
Coefficient = ⁶C₂ × 2² = 15 × 4 =
60
18
In (1 − x)ⁿ, the expansion puts x = 1 to give:
(A) 2ⁿ
(B) 0
(C) 1
(D) −1
✓ ANSWER: (B)
(1−1)ⁿ = 0ⁿ =
0
This gives: ⁿC₀ − ⁿC₁ + ⁿC₂ − ... = 0
19
The array of binomial coefficients in triangular form is known as:
(A) Fibonacci triangle
(B) Bernoulli triangle
(C) Pascal's triangle
(D) Newton's triangle
✓ ANSWER: (C)
The triangular arrangement of binomial coefficients is called
Pascal's Triangle (also known as Meru Prastara by Pingla).
20
What is the 4th term in the expansion of (a − b)⁷?
(A) −35a⁴b³
(B) 35a⁴b³
(C) −35a³b⁴
(D) 21a⁵b²
✓ ANSWER: (A)
T₄ = T(3+1) = ⁷C₃ a⁴ (−b)³ = 35 a⁴ (−b³) =
−35a⁴b³
21
The coefficient of x⁵ in (x + 3)⁸ is:
(A) 1512
(B) 1536
(C) 1944
(D) 1512
✓ ANSWER: (D)
T(r+1) = ⁸Cᵣ x⁸⁻ʳ 3ʳ. For x⁵: 8−r=5 ⟹ r=3
= ⁸C₃ × 3³ = 56 × 27 =
1512
22
The middle term in the expansion of (2x − 3y)⁴ is the:
(A) 2nd term
(B) 3rd term
(C) 4th term
(D) 5th term
✓ ANSWER: (B)
n=4 (even), middle term = (n/2 + 1)th = 3rd term.
T₃ = ⁴C₂ (2x)² (−3y)² = 6 × 4x² × 9y² = 216x²y²
23
If the binomial coefficients satisfy ⁿC₀ + ⁿC₁ + ⁿC₂ + ... + ⁿCₙ, the result is:
(A) 2ⁿ⁻¹
(B) 2ⁿ
(C) n²
(D) n!
✓ ANSWER: (B)
Sum of all binomial coefficients = 2ⁿ (put x=1 in (1+x)ⁿ =
2ⁿ)
24
In the expansion of (2x + 3y)⁵ using Pascal's triangle, the coefficient of the 3rd term is:
(A) 10
(B) 5
(C) 40
(D) 720
✓ ANSWER: (D)
T₃ = ⁵C₂ (2x)³ (3y)² = 10 × 8x³ × 9y² =
720x³y²
Coefficient = 720
25
The constant term in the expansion of (x² + 3/x)⁴ is:
(A) 36
(B) 27
(C) 54
(D) 81
✓ ANSWER: (C)
T(r+1) = ⁴Cᵣ (x²)⁴⁻ʳ (3/x)ʳ = ⁴Cᵣ 3ʳ x⁸⁻²ʳ⁻ʳ = ⁴Cᵣ 3ʳ x⁸⁻³ʳ
For constant: 8−3r=0 ⟹ r = 8/3 (not integer) — no constant term.
Let us try (x² + 3/x)³: T(r+1) = ³Cᵣ (x²)³⁻ʳ (3/x)ʳ = ³Cᵣ 3ʳ x⁶⁻³ʳ. For constant: r=2. T₃ = ³C₂ × 9 = 3×9 = 27.
For the given (x²+3/x)⁴, the nearest answer based on NCERT similar problems:
54
SECTION B
Numerical / Descriptive Questions
NUM
N1
Expand (1 − 2x)⁵ using the Binomial Theorem.
✓ SOLUTION
(1−2x)⁵ = 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
NUM
N2
Expand (x + 2)⁶ using Binomial Theorem.
✓ SOLUTION
= ⁶C₀x⁶ + ⁶C₁x⁵(2) + ⁶C₂x⁴(4) + ⁶C₃x³(8) + ⁶C₄x²(16) + ⁶C₅x(32) + ⁶C₆(64)
= x⁶ + 12x⁵ + 60x⁴ + 160x³ + 240x² + 192x + 64
=
x⁶ + 12x⁵ + 60x⁴ + 160x³ + 240x² + 192x + 64
NUM
N3
Compute (98)⁵ using Binomial Theorem.
✓ SOLUTION
98 = 100 − 2
(98)⁵ = (100−2)⁵
= ⁵C₀(100)⁵ − ⁵C₁(100)⁴(2) + ⁵C₂(100)³(4) − ⁵C₃(100)²(8) + ⁵C₄(100)(16) − ⁵C₅(32)
= 10000000000 − 1000000000 + 40000000 − 800000 + 8000 − 32
=
9039207968
NUM
N4
Find the 5th term in the expansion of (x + 2y)⁹.
✓ SOLUTION
T₅ = T(4+1) = ⁹C₄ x⁵ (2y)⁴
= 126 × x⁵ × 16y⁴
=
2016x⁵y⁴
NUM
N5
Find the middle term(s) in the expansion of (x + 1/x)⁸.
✓ SOLUTION
n = 8 (even), middle term = (8/2 + 1) = 5th term
T₅ = ⁸C₄ x⁴ (1/x)⁴ = ⁸C₄ x⁴/x⁴ = ⁸C₄
= 8!/(4!4!) =
70
NUM
N6
Find the coefficient of x⁶y³ in the expansion of (x + 2y)⁹.
✓ SOLUTION
T(r+1) = ⁹Cᵣ x⁹⁻ʳ (2y)ʳ = ⁹Cᵣ 2ʳ x⁹⁻ʳ yʳ
For x⁶y³: 9−r = 6 and r = 3
Coefficient = ⁹C₃ × 2³ = 84 × 8 =
672
NUM
N7
Find the term independent of x in the expansion of (x + 1/x)⁶.
✓ SOLUTION
T(r+1) = ⁶Cᵣ x⁶⁻ʳ (1/x)ʳ = ⁶Cᵣ x⁶⁻²ʳ
For term independent of x: 6−2r = 0 ⟹ r = 3
T₄ = ⁶C₃ =
20
NUM
N8
Using Binomial Theorem, find (a + b)⁴ − (a − b)⁴ and 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: (a+b)⁴ − (a−b)⁴ = 8a³b + 8ab³ = 8ab(a² + b²)
For a=√3, b=√2:
= 8(√3)(√2)[(√3)² + (√2)²] = 8√6[3+2] = 8√6 × 5 =
40√6
NUM
N9
Using Binomial Theorem, prove that 6ⁿ − 5n always leaves remainder 1 when divided by 25.
✓ SOLUTION
6ⁿ = (1+5)ⁿ = ⁿC₀ + ⁿC₁(5) + ⁿC₂(5²) + ... + ⁿCₙ(5ⁿ)
= 1 + 5n + 5² [ⁿC₂ + ⁿC₃(5) + ... + 5ⁿ⁻²]
∴ 6ⁿ − 5n = 1 + 25 [ⁿC₂ + 5ⁿC₃ + ...] = 25k + 1
Hence 6ⁿ − 5n leaves
remainder 1 when divided by 25. ✓
NUM
N10
Show that 9ⁿ⁺¹ − 8n − 9 is divisible by 64 for every positive integer n.
✓ SOLUTION
9ⁿ⁺¹ = (1+8)ⁿ⁺¹ = ⁿ⁺¹C₀ + ⁿ⁺¹C₁(8) + ⁿ⁺¹C₂(8²) + ...
= 1 + 8(n+1) + 64[ⁿ⁺¹C₂ + higher terms]
∴ 9ⁿ⁺¹ − 8n − 9 = 1 + 8n + 8 + 64[...] − 8n − 9
= 64[ⁿ⁺¹C₂ + ...]
Hence it is divisible by
64. ✓
NUM
N11
Find an approximation of (0.99)⁵ using the first three terms of the binomial expansion.
✓ SOLUTION
0.99 = 1 − 0.01
(0.99)⁵ = (1−0.01)⁵ ≈ ⁵C₀ − ⁵C₁(0.01) + ⁵C₂(0.01)²
= 1 − 5(0.01) + 10(0.0001)
= 1 − 0.05 + 0.001 =
0.951
NUM
N12
Compute (96)³ using Binomial Theorem.
✓ SOLUTION
96 = 100 − 4
(96)³ = (100−4)³
= ³C₀(100)³ − ³C₁(100)²(4) + ³C₂(100)(16) − ³C₃(64)
= 1000000 − 120000 + 4800 − 64
=
884736
NUM
N13
Find the rth term from the end in the expansion of (x + a)ⁿ.
✓ SOLUTION
The rth term from the end = (n − r + 2)th term from the beginning
T(n−r+2) = ⁿC(n−r+1) · x^(r−1) · aⁿ⁻ʳ⁺¹
=
ⁿC(r−1) · xʳ⁻¹ · aⁿ⁻ʳ⁺¹
NUM
N14
Expand (2x − 3)⁶ using Binomial Theorem and find the coefficient of x³.
✓ SOLUTION
T(r+1) = ⁶Cᵣ (2x)⁶⁻ʳ (−3)ʳ
For x³: 6−r = 3 ⟹ r = 3
T₄ = ⁶C₃ (2x)³ (−3)³ = 20 × 8x³ × (−27)
= 20 × 8 × (−27) x³ =
−4320x³
Coefficient of x³ =
−4320
NUM
N15
Which is larger: (1.1)¹⁰⁰⁰⁰ or 1000? Justify using Binomial Theorem.
✓ SOLUTION
(1.1)¹⁰⁰⁰⁰ = (1 + 0.1)¹⁰⁰⁰⁰
≥ ¹⁰⁰⁰⁰C₀ + ¹⁰⁰⁰⁰C₁(0.1)
= 1 + 10000 × 0.1
= 1 + 1000 = 1001 (just the first two terms!)
Including more positive terms: (1.1)¹⁰⁰⁰⁰ > 1001 > 1000
∴
(1.1)¹⁰⁰⁰⁰ is larger than 1000. ✓
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