Exercise 816Combinatorics › Binomial coefficientsL1Added 3 Nov 2026

Weighted sum of binomial coefficients

#weighted sum#binomial#derivative

Statement

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

Hint

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

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Derivative. $\frac{d}{dx}(1+x)^n=n(1+x)^{n-1}=\sum_{k=0}^{n}k\binom{n}{k}x^{k-1}$ .

At $x=1$ . $n\cdot 2^{n-1}=\sum_{k=0}^{n}k\binom{n}{k}$ .

Conclusion. $\sum_k k\binom{n}{k}=n\cdot 2^{n-1}$ .