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📐 Mathematics  ·  Binomial Theorem  ·  JEE

If (1 + x)ⁿ = C₀ + C₁x + C₂x² + ..., then C₀ + 2C₁ + 3C₂ + ... + (n+1)Cₙ =

Answer: (n+2)2ⁿ⁻¹.

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

Correct answer: A. (n+2)2ⁿ⁻¹

Explanation: Sum = (n+1)2<sup>n-1</sup> + 2<sup>n-1</sup>. Actually: C<sub>0</sub>+2C1+...+(n+1)Cn = sum(r=0 to n) (r+1)Cr = sum(r+1)nCr = n*2<sup>n-1</sup>+2<sup>n</sup> = (n+2)*2<sup>n-1</sup>.

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

Read the full Binomial Theorem notes →