Find the value of ∑k=1nk(nk)?

From the binomial theorem, we have
${\displaystyle {(1+x)}^n=\sum _{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)x^k}$ (1)
Differentiating (1) reveals
${\displaystyle n{(1+x)}^{n-1}=\sum _{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)k{x}^{k-1}}$ (2)
Setting ${\displaystyle x=1}$ in (2) yields
${\displaystyle n{2}^{n-1}=\sum _{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)k}$
And we are done!

We have
${\displaystyle \sum _{k=0}^{n}k\left(\begin{array}{c}n\\ k\end{array}\right)=\sum _{k=1}^{n}k\left(\begin{array}{c}n\\ k\end{array}\right)}$
${\displaystyle n=\sum _{k=1}^n\frac{(n-1)!}{(k-1)!(n-k)!}}$
${\displaystyle n\sum _{l=0}^{n-1}\frac{(n-1)!}{l((n-1)-l)!}}$
${\displaystyle =n\sum _{l=0}^{n-1}\left(\begin{array}{c}n-1\\ l\end{array}\right)}$
${\displaystyle =n{2}^{n-1}}$

Proof without derivatives: $\sum _{k=1}^{n}k((n),(k))=\sum _{k=1}^{n}k\frac{n!}{(n-k)!k!}$ $=\sum _{k=1}^n\frac{n!}{(n-k)!(k-1)!}$ $=\sum _{k=1}^n\frac{n(n-1)!}{(n-k)!(k-1)!}$ $=\sum _{k=1}^{n}n((n-1),(k-1))$ $=n\sum _{k=0}^{n-1}((n-1),(k))=n{2}^{n-1}$ Alternate proof via probability theory: Toss a fair coin n times, find the expected no of heads. Let N be the random variable denoting the number of heads. Then $E[N]=n/2$ because N is the sum of n bernoulli random variables with probability 1/2. But we also know that N has a binomial distribution. Hence $E[N]=\sum _{k=1}^{n}k((n),(k))2^{-n}$ Rearrange to get your answer.