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.
Just change the index $s=k-1$
$$\sum_{k=1}^{n}k {n \choose k} = n\sum_{k=1}^{n} {n-1\choose k-1}= n\sum_{s=0}^{n-1} {n-1\choose s} =n2^{n-1}$$
\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*}
One possible proof:
Because $$(1+x)^n = \sum_{k=0}^n {n \choose k} x^k,$$ then by taking derivative, $$n(1+x)^{n-1} = \sum_{k=1}^n k{n \choose k} x^{k-1}. $$ Let $x=1$, we obtain that $$n2^{n-1} = \sum_{k=1}^n k {n \choose k}.$$
Another way to do this is to differentiate the identity $$(1 + x)^n = \sum_{k=0}^n \binom{n}{k}x^k $$ and then put $x = 1$.
