Answer: n·2ⁿ⁻¹.
- A n·2ⁿ
- B 2ⁿ
- C (n+1)·2ⁿ⁻¹
- D n·2ⁿ⁻¹
Correct answer: D. n·2ⁿ⁻¹
Explanation: Technique: differentiate (1+x)n=ΣnCr·xʳ to get n(1+x)n−1=Σr·nCr·xr−1. Set x=1: n·2n−1=Σr·nCr. Alternatively use identity r·nCr=n·(n−1)C(r−1) and sum. Answer: n·2n−1.
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