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The value of (nC<sub>1</sub>/nC<sub>0</sub>) + 2(nC<sub>2</sub>/nC<sub>1</sub>) + 3(nC<sub>3</sub>/nC<sub>2</sub>) + ... + n(nCn/nCn-1) is:

Answer: n(n+1)/2.

  • A n(n+1)/2
  • B n(n-1)/2
  • C n(n+1)
  • D n²(n+1)/2

Correct answer: A. n(n+1)/2

Explanation: Simplify each ratio: nCr/nC(r−1)=(n−r+1)/r. So the r-th term is r·(n−r+1)/r=n−r+1. Sum=Σ(r=1 to n)(n−r+1)=n+(n−1)+…+1=n(n+1)/2. Uses consecutive-ratio reduction technique.

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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