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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
    56
    Correct
  2. B
    12
  3. C
    15
  4. D
    35

Correct answer

56

Explanation

Since 1+2x+3x2+⋯=(1−x)−21+2x+3x^2+\cdots=(1-x)^{-2}, its square is (1−x)−4(1-x)^{-4}. The coefficient of x5x^5 is (5+4−14−1)=(83)=<strong>56</strong>\binom{5+4-1}{4-1}=\binom83=<strong>56</strong>.

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