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?

Notice that $\displaystyle S = \sum_{k=0}^{n} k \binom {n} {k}  = \sum_{k=0}^{n}(n-k)\binom {n} {n-k} = n\sum_{k=0}^{n}\binom {n} {n-k}-S$
so, as $\displaystyle \binom {n} {n-k}=\binom {n} {k}$ we have $\displaystyle 2S  = n\sum_{k=0}^{n}\binom {n} {k} = n2^n$ and so $\displaystyle S = n2^{n-1}.$
