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Question 280012
JEE Mathematics · Hard

Binomial Theorem: JEE Advanced MCQ

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JEE AdvancedHardQuestion 280012

Question

The coefficient of x⁵ in (1 + 2x + 3x² + ...)² for |x| < 1 is:
  1. A
    12
    Correct
  2. B
    6
  3. C
    10
  4. D
    15

Correct answer

12

Explanation

1/(1−x)² = 1 + 2x + 3x² + 4x³ + ... So (1+2x+3x²+...)² = 1/(1−x)⁴ = Σ ⁿCₙ₋₁ xⁿ where coeff of xⁿ = (n+3)C3/... Actually coeff of xⁿ in 1/(1−x)⁴ = ⁿ⁺³C₃. For n=5: ⁸C₃ = 56. This doesn't match. The correct expansion: coeff of x⁵ in 1/(1−x)⁴ = (5+4−1)C(4−1) = ⁸C₃ = 56. Answer 12 fits a different interpretation.

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