Proof by induction $\sum_{k=1}^{n}$ $k \binom{n}{k}$ $= n\cdot 2^{n-1}$ for each natural number $n$ Prove by induction that $\sum_{k=1}^{n}$ $k \binom{n}{k}$ $= n\cdot 2^{n-1}$ for each natural number $n$.

Observe that: $\binom{n+1}{k} = \binom{n}{k} + \binom{n}{k-1}$. Thus: $\displaystyle \sum_{k=1}^{n+1} k\binom{n+1}{k} = \displaystyle \sum_{k=1}^{n+1} k\binom{n}{k} + \displaystyle \sum_{k=1}^{n+1} k\binom{n}{k-1} = \displaystyle \sum_{k=1}^n k\binom{n}{k} + \displaystyle \sum_{k=0}^n (k+1)\binom{n}{k} = 2\displaystyle \sum_{k=1}^n k\binom{n}{k} + \displaystyle \sum_{k=0}^n \binom{n}{k} = 2\cdot n2^{n-1} + 2^n = (n+1)2^n$
