Binomial coefficient identity $\sum_{k=1}^n k {n \choose k } = n\cdot 2^{n-1}$

Question

I'm having a bit of problems proving the following: $$\sum_{k=1}^n k {n \choose k } = n\cdot 2^{n-1}$$

I always seem to get to the line: $2^{n-1} + 1 = 2^n$ which I know is untrue.

Could anyone help me prove this?

Answer 1

Consider the identity $(1+x)^n=\sum_{i=0}^n\binom{n}{i}x^i$. Derive both sides with respect to $x$ and you obtain $$n(1+x)^{n-1}=\sum_{i=0}^n i\binom{n}{i}x^{i-1}.$$ You can get your result setting $x=1$.

Answer 2

Start from $(1+x)^n=\displaystyle\sum_{k=0}^{n}\dbinom nk x^k$ and derive both members, then set $x=1$.