Answer: (2n)Cn.
- A (2n)Cn
- B 2ⁿ
- C n!
- D nC(n/2)
Correct answer: A. (2n)Cn
Explanation: Recognise: 1=nC<sub>0</sub>=nC<sub>0</sub>², nC<sub>1</sub>=nC<sub>1</sub>² only if nC<sub>1</sub>=1 (n=1). In general the series is nC<sub>0</sub>²+nC<sub>1</sub>²+…+nCn². By Vandermonde identity this equals coeff of xⁿ in (1+x)<sup>2n</sup>=(2n)Cn. Answer: (2n)Cn.
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