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📐 Mathematics  ·  Binomial Theorem  ·  JEE

If (1 + x)ⁿ = C₀ + C₁x + ... + Cₙxⁿ, then C₀ + 2C₁ + 3C₂ + ... + (n+1)Cₙ equals:

Answer: 2 n−1 (n + 2).

  • A 2<sup>n−1</sup>(n + 2)
  • B 2ⁿ(n + 1)
  • C n·2ⁿ
  • D (n + 2)2ⁿ

Correct answer: A. 2<sup>n−1</sup>(n + 2)

Explanation: Σ(r+1)C(n,r) = n·2<sup>n−1</sup> + 2ⁿ = 2<sup>n−1</sup>(n + 2).

Pascal's Triangle: Binomial Coefficients11112113311464115101051n=0n=1n=2n=3n=4n=5Each entry = sum of the two entries diagonally above it (Pascal's identity); row n gives nC0...nCn

Row n of Pascal's triangle gives the coefficients nC0, nC1, ..., nCn for the expansion of (a+b)ⁿ; each number is the sum of the two numbers diagonally above it, a direct visual proof of the identity nCr = (n-1)C(r-1) + (n-1)Cr.

Concept context

Expansion of (a+b) n , general term, middle term, binomial coefficients, and greatest term

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