Find the value of $\sum_{k=1}^{n} k \binom {n} {k}$ I was assigned the following problem: find the value of $$\sum_{k=1}^{n} k \binom {n} {k}$$ by using the derivative of $(1+x)^n$, but I'm basically clueless. Can anyone give me a hint?

Imagine tossing a fair coin $n$ times. Then the mean number of heads is
$$\sum_0^n k\binom{n}{k}\frac{1}{2^n}.\tag{1}$$
We compute the mean another way. Let random variable $X_i$ be $1$ if we get a head on the $i$-th toss, and $0$ otherwise. Then the number $Y$ of heads is given by 
$$Y=X_1+X_2+\cdots +X_n,$$
and therefore by the linearity of expectation we have
$$E(Y)=E(X_1)+E(X_2)+\cdots +E(X_n).$$
Since $X_i=1$ with probability $\frac{1}{2}$, we have
$$E(Y)=\frac{n}{2}.\tag{2}$$
Now comparison of (1) and (2) gives the result.
Remark: This is an example of a Mean Proof. 
