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

Question

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

Answer 1

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

Answer 2

Here is a way to derive this result. By the binomial theorem, $$(1+x)^n =\sum^{n}_{k=0}\binom{n}{k}x^k$$ Differentiate both sides. $$n(1+x)^{n-1} =\sum^{n}_{k=0} k\binom{n}{k}x^{k-1}$$ Substitute $x=1$ $$n2^{n-1} =\sum^{n}_{k=0} k\binom{n}{k} =\sum^{n}_{k=1} k\binom{n}{k} $$