Exercise 1609Combinatorics › Counting principlesOlympiadAdded 3 Nov 2026

Alternating sum identity

#alternating sum#binomial#identity

Statement

Show $\sum_{k=1}^{n}k\binom{n}{k}(-1)^{k+1}=0$ for $n\ge2$ .

Hint

Differentiate $(1-x)^n$ and set $x=1$ .

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Derivative. $\frac{d}{dx}(1-x)^n=-n(1-x)^{n-1}=\sum_k k\binom{n}{k}(-x)^{k-1}(-1)$ .

Set $x=1$ . LHS $=0$ for $n\ge2$ ; RHS $=-\sum_k k\binom{n}{k}(-1)^{k-1}$ .

Conclusion. $\sum_k k\binom{n}{k}(-1)^{k-1}=0$ .