Find the value of $\sum_{k=1}^{n}k\binom{n}{k}$?

From the binomial theorem, we have

$$(1+x)^n=\sum_{k=0}^n\binom{n}{k}x^k\tag 1$$

Differentiating $(1)$ reveals

$$n(1+x)^{n-1}=\sum_{k=0}^n\binom{n}{k}kx^{k-1}\tag2$$

Setting $x=1$ in $(2)$ yields

$$n2^{n-1}=\sum_{k=0}^n\binom{n}{k}k$$

And we are done!

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Interestingly, I showed in THIS ANSWER, that for $m<n$, we have $$\sum_{k=0}^n\binom{n}{k}(-1)^k k^m=0$$

We have \begin{align*} \sum_{k=0}^nk{n\choose k} &=\sum_{k=1}^nk{n\choose k}\\ &=n\sum_{k=1}^n\frac{(n-1)!}{(k-1)!(n-k)!}\\ &=n\sum_{\ell=0}^{n-1}\frac{(n-1)!}{\ell!((n-1)-\ell)!} \tag{by taking $\ell=k-1$}\\ &=n\sum_{\ell=0}^{n-1}{n-1\choose\ell}\\ &=n2^{n-1}. \end{align*}

There is also a combinatorial argument:

Suppose you have a room of $n$ people and want to select a committee of $k$ of them, where one member is the chairperson. There are $k\binom{n}{k}$ ways to do this. Your sum represents the total number of ways to select such a committee of any size (from $1$ to $n$) with a chairperson.

How else can we think of this? Instead, first pick the committee chairperson. There are $n$ ways to do this. Then, go to each of the remaining $n-1$ people and decide if they should be in the committee. There are $2^{n-1}$ ways to do this. Note that we can create any committee/chairperson team this way, as before. Hence your sum is equal to $n 2^{n-1}$.