Exercise 1548Combinatorics › Binomial coefficientsOlympiadAdded 3 Nov 2026

Weighted power sum

#weighted sum#binomial#derivative

Statement

Show that $\sum_{k=0}^{n}k^2\binom{n}{k}=n(n+1)2^{n-2}$ .

Hint

Differentiate $(1+x)^n$ twice, then multiply by $x$ .

Show solutionHide solution

First derivative. $\sum_k k\binom{n}{k}x^{k-1}=n(1+x)^{n-1}$ .

Multiply by $x$ . $\sum_k k\binom{n}{k}x^k=nx(1+x)^{n-1}$ .

Differentiate. $\sum_k k^2\binom{n}{k}x^{k-1}=n(1+x)^{n-1}+n(n-1)x(1+x)^{n-2}$ .

At $x=1$ . $\sum_k k^2\binom{n}{k}=n\cdot 2^{n-1}+n(n-1)2^{n-2}=2^{n-2}(2n+n^2-n)=2^{n-2}(n^2+n)=n(n+1)2^{n-2}$ .

Conclusion. The identity holds.