How to proof $\sum_{k=1}^n k {n \choose k }=n 2^{n-1}$

Question

I have tried trying to find a pattern but i don't believe that the right way. If you help me it would be great.

Answer 1

Recall that the binomial formula states that $$(x+y)^n=\sum_{k=0}^{n}\binom{n}{k}x^{k}y^{n-k} $$ Differentiating with respect to $x$ yields $$n(x+y)^{n-1}=\sum_{k=0}^{n}\binom{n}{k}kx^{k-1}y^{n-k}$$

Then taking $x=y=1$, the result follows.