TITLE: How to prove $\sum_{k=1}^{n}k\binom{n}{k} = n2^{n-1}$
QUESTION [5 upvotes]: $\sum_{k=1}^{n}k\binom{n}{k} = n2^{n-1}$
I have tried both induction and transforming both sides to get equality but no luck
I know that 
$\sum_{k=1}^{n}\binom{n}{k} = 2^{n}-1$
and $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$
p.s: I couldn't find something similar of this kind anywhere. But I thinks it's something that is related to Pascals rule.
Please point out little hints, thank you

REPLY [6 votes]: You can use $\binom{n}{k} = \binom{n}{n-k}$:
$$
\begin{align*}
2\sum_{k=0}^n k\binom{n}{k} &= \sum_{k=0}^n k\binom{n}{k} + \sum_{k=0}^n k\binom{n}{n-k} \\ &= \sum_{k=0}^n k\binom{n}{k} + \sum_{k=0}^n (n-k)\binom{n}{k} \\ &= \sum_{k=0}^n [k+(n-k)]\binom{n}{k} \\ &= \sum_{k=0}^n n\binom{n}{k} \\ &= n2^n.
\end{align*}
$$
In the first line we used $\binom{n}{k} = \binom{n}{n-k}$, in the second we used the substitution $k = n-k$ in the second sum.