TITLE: Proving equality - a sum including binomial coefficient $\sum_{k=1}^{n}k{n \choose k}2^{n-k}=n3^{n-1}$
QUESTION [5 upvotes]: I want to prove the following equality:
$$\displaystyle\sum_{k=1}^{n}k{n \choose k}2^{n-k}=n3^{n-1}$$
So I had an idea to use $((1+x)^n)'=n(1+x)^{n-1}$
So I could just use the binomial theorem and let $$x=2 \implies (n3^{n-1})$$ and then modify the sum into that one on the left side.
So I need to prove that :
$$\displaystyle\sum_{k=1}^{n}k{n \choose k}2^{n-k}=n(1+x)^{n-1}$$ if$$x=2$$
Any help would be appreciated.

REPLY [3 votes]: We can do it without derivatives using the identity $\color{red}{\binom{n+1}{k+1}={n+1\over k+1}\binom{n}{k}}$, then using the change of variable $\color{blue}{k=r+1}$ and lastly using the binomial theorem.
$$\begin{align}
S&=\sum_{k=1}^{n}k{n \choose k}2^{n-k}\color{red}=\sum_{k=1}^{n}n{n-1 \choose k-1}2^{n-k}\color{blue}=n\sum_{r=0}^{n-1}{n-1\choose r}2^{(n-1)-r}=n(1+2)^{n-1}=n3^{n-1}
\end{align}$$