Prove that $\sum_{k=1}^{n}k\binom{n}{k}=n\cdot2^{n-1}$

Question

I want to prove the following:
$$\sum_{k=1}^{n}k\binom{n}{k}=n\cdot2^{n-1}$$ what I did is(use binominal):
$$\binom{n}{k}X^k\cdot 1^{n-k} = (X+1)^n$$ $$k\binom{n}{k}X^k\cdot 1^{n-k} = k(X+1)^k-1$$ now I replace $k$ by $n$ and insert $X=1$ , I will get now $n\cdot2^{n-1}$
what you are suggesting? there is another way to do that?
thanks!

Answer 1

note $$k\binom{n}{k}=n\binom{n-1}{k-1}$$ so $$\sum_{k=1}^{n}k\binom{n}{k}=n\sum_{k=1}^{n}\binom{n-1}{k-1}=n(1+1)^{n-1}=n2^{n-1}$$

Answer 2

A combinatorial proof:

Suppose you want to choose a subset of $\{1, 2, \dots, n\}$ and a ''special element'' from that subset. You can imagine it like picking out players for a match and one captain out of them.

RHS: You first pick a captain (you can do it in $n$ ways), and from remaining players you pick a ''team'' (choosing a subset of $n-1$-element set, you can do it in $2^{n-1}$ ways.

LHS: Suppose you want to pick $k$ players (including a captain). You choose $k$ players out of $n$ (you can do it in $n \choose k$ ways, and then, out of them, you choose a captain (you can do it in $k$ ways). Sum over $k \in \{1, 2, \dots, n\}$.