TITLE: Sum of $k {n \choose k}$ is $n2^{n-1}$
QUESTION [15 upvotes]: Proof that $\suṃ̣_{k=1}^{n}k {n \choose k}$ for $n \in \mathbb N$ is equal to $n2^{n-1}$.
As a hint I got that $k {n \choose k} = n {n-1\choose k-1} $.
I tried solving this by induction but, in the inductive step I'm not arriving to the correct result.

REPLY [7 votes]: \begin{align*}  &A = \sum_{k\geq1}^{n}k\cdot \binom{n}{k} \\ \end{align*}

$A $ is the no of ways to form a committee of $k \geq 1$ people out of $n$ available individuals and select one head for the selected committee.
Empty committee is not possible. Therefore we select the head first in $n$ ways and then any subset out of $(n-1)$ remaining people in $2^{n-1} $ ways.

\begin{align*}  &A =n\cdot2^{n-1} \\  \end{align*}