Exercise 412Combinatorics › Binomial coefficientsHigh schoolAdded 3 Nov 2026

Sum over odd and even sizes

#sum parity#binomial#subsets

Statement

Show that $\sum_{k\text{ odd}}k\binom{n}{k}=n\cdot 2^{n-2}$ for $n\ge2$ .

Hint

Use symmetric sums or generating functions.

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Approach. Total $\sum_k k\binom{n}{k}=n\cdot 2^{n-1}$ . Even and odd differ by $\sum_k(-1)^kk\binom{n}{k}$ .

Alternating sum. $\sum_k(-1)^kk\binom{n}{k}=0$ for $n\ge2$ (derivative argument).

Compute. Even sum = odd sum = $\frac{n\cdot2^{n-1}}{2}=n\cdot2^{n-2}$ .

Conclusion. $\sum_{k\text{ odd}}k\binom{n}{k}=n\cdot2^{n-2}$ .