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Binomial Theorem – Class 11 Mathematics notes, important questions and MCQs

NCERT Class 11 Mathematics, chapter 7. Notes, key terms, important questions with answers and MCQs.

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Chapter overview

The binomial theorem expands (a + b)ⁿ for a positive integer n using binomial coefficients from Pascal's triangle.

Expansion

(a + b)ⁿ = Σ ⁿCᵣ aⁿ⁻ʳ bʳ, r = 0 to n; there are n + 1 terms.
General term Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ.
Sum of coefficients = 2ⁿ (put a = b = 1).

Middle term and uses

n even: one middle term T₍ₙ/₂₎₊₁; n odd: two middle terms.
Coefficients are symmetric: ⁿCᵣ = ⁿCₙ₋ᵣ.
Used for approximations like (1.01)⁵ and divisibility proofs.

Key terms

Binomial coefficient: ⁿCᵣ, the coefficient in the expansion
Pascal's triangle: triangular array where each number is the sum of the two above
General term: Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ, the (r + 1)th term in the expansion of (a + b)ⁿ

Formulas

Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ
Sum of coefficients = 2ⁿ

Important questions with answers

Expand (x + 1/x)⁴.2 marks

x⁴ + 4x² + 6 + 4/x² + 1/x⁴.

Using the binomial theorem, find (99)⁴.3 marks

(100 − 1)⁴ = 100⁴ − 4 × 100³ + 6 × 100² − 4 × 100 + 1 = 100,000,000 − 4,000,000 + 60,000 − 400 + 1 = 96,059,601.

Find the middle terms of (3x − x³/6)⁷, and show that 6ⁿ − 5n leaves remainder 1 when divided by 25.5 marks

n = 7 is odd, so there are two middle terms, T₄ and T₅. T₄ = ⁷C₃ (3x)⁴ (−x³/6)³ = 35 × 81x⁴ × (−x⁹/216) = −105x¹³/8. T₅ = ⁷C₄ (3x)³ (−x³/6)⁴ = 35 × 27x³ × x¹²/1296 = 35x¹⁵/48. For the second part: 6ⁿ = (1 + 5)ⁿ = 1 + 5n + ⁿC₂ 5² + ⁿC₃ 5³ + …, so 6ⁿ − 5n = 1 + 25(ⁿC₂ + ⁿC₃ 5 + …), which is 1 more than a multiple of 25.

Quick revision

n + 1 terms.
Coefficients from Pascal's triangle.
Sum of coefficients 2ⁿ.

Exam tips

For the term independent of x, set the power of x in Tᵣ₊₁ to zero.
To find a coefficient, write the general term, simplify the power of x, and equate it to the required power.

Practice MCQs

Number of terms in (a + b)⁷ is
1. 7
2. 8
3. 6
4. 14
Show answer
Answer: (b) 8. n + 1.
Sum of coefficients in (1 + x)⁵ is
1. 5
2. 10
3. 25
4. 32
Show answer
Answer: (d) 32. 2⁵.
Coefficient of x² in (1 + x)⁵ is
1. 5
2. 10
3. 20
4. 1
Show answer
Answer: (b) 10. ⁵C₂.
(x + y)⁴ has middle term
1. 6x²y²
2. 4x³y
3. 4xy³
4. x²y²
Show answer
Answer: (a) 6x²y². T₃ = ⁴C₂x²y².
The 4th term of (x + 2)⁶ is
1. 20x³
2. 160x³
3. 80x³
4. 240x²
Show answer
Answer: (b) 160x³. ⁶C₃ x³ 2³.
Term independent of x in (x + 1/x)⁶ is
1. 20
2. 15
3. 6
4. 1
Show answer
Answer: (a) 20. ⁶C₃.
(1.1)⁵ is closest to
1. 1.5
2. 1.21
3. 1.61
4. 2
Show answer
Answer: (c) 1.61. 1 + 0.5 + 0.1 + 0.01 ≈ 1.61.
The coefficient of x⁵ in (1 + 2x)⁶ is
1. 160
2. 32
3. 6
4. 192
Show answer
Answer: (d) 192. ⁶C₅ × 2⁵.

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Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.