Sum with $k^2\binom{n}{k}$

#weighted sum#square#binomial

Statement

Compute $\sum_{k=0}^{n}k^2\binom{n}{k}$ .

Hint

Differentiate twice.

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

See "Weighted power sum" in this category for the proof.

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