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📐 Mathematics  ·  Binomial Theorem  ·  JEE

The sum C₀² + C₁² + C₂² + ... + Cₙ² (where Cᵣ = nCr) equals:

Answer: (2n)Cn.

  • A 2ⁿ
  • B (2n)Cn
  • C 2nCn
  • D n!

Correct answer: B. (2n)Cn

Explanation: Vandermonde identity: Σ(r=0 to n)(nCr)²=coeff of xⁿ in (1+x)<sup>n</sup>·(1+x)<sup>n</sup>=(1+x)<sup>2n</sup>, which equals (2n)Cn. Since nCr=nC(n−r), this is equivalent to coeff of xⁿ in (1+x)<sup>2n</sup>. Answer: (2n)Cn.

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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