How Find this sum $$\sum_{k=0}^{n}k\binom{n+k}{k}2^k$$
My idea: since $$\binom{n+k}{k}k=\dfrac{(n+k)!}{n!(k-1)!}$$ and I have other idea: Consider $$f(x)=\sum_{k=0}^{n}\binom{n+k}{k}x^k$$ then $$f'(x)=\sum_{k=0}^{n}k\binom{n+k}{k}x^{k-1}$$
But $f$ I can't find the closed form.($f(1/2)=2^n$,can see this link)
I find sometimes to find this
Proof of the identity $2^n = \sum\limits_{k=0}^n 2^{-k} \binom{n+k}{k}$ or can see Prove the following relation:
(But my question is different this )
This wolf can't :http://www.wolframalpha.com/input/?i=%5Csum_%7Bk%3D1%7D%5E%7Bn%7Dk2%5Ek%5Cbinom%7Bn%2Bk%7D%7Bk%7D
Thank you