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📐 Mathematics  ·  Binomial Theorem  ·  JEE

Find the value of C₀ + C₁/2 + C₂/3 + ... + Cₙ/(n+1) where Cᵣ = nCr.

Answer: (2ⁿ⁺¹-1)/(n+1).

  • A (2ⁿ+1)/(n+1)
  • B 2ⁿ/(n+1)
  • C (2ⁿ⁺¹-1)/(n+1)
  • D 2ⁿ⁻¹/(n+1)

Correct answer: C. (2ⁿ⁺¹-1)/(n+1)

Explanation: Integrate (1+x)<sup>n</sup> from 0 to 1: [(1+x)<sup>n+1</sup>/(n+1)] from 0 to 1 = (2<sup>n+1</sup>-1)/(n+1). LHS integral = C<sub>0</sub> + C<sub>1</sub>/2 + C<sub>2</sub>/3 + ... + Cn/(n+1). So answer = (2<sup>n+1</sup>-1)/(n+1).

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 →