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

  • $\begingroup$ You can have a closed form in terms of hypergeometric function. $\endgroup$ Feb 12, 2014 at 7:23
  • $\begingroup$ @MhenniBenghorbal,why? Thank you $\endgroup$
    – user94270
    Feb 12, 2014 at 7:34
  • $\begingroup$ It is really new. I've seen $2^n = \sum\limits_{k=0}^n 2^{-k} \binom{n+k}{k}$, but you cannot use this technique. $\endgroup$
    – Bobby
    Feb 12, 2014 at 7:35

1 Answer 1


As said by Mhenni Benghorbal $$S(m)=\sum_{k=0}^{n}k\binom{n+k}{k}m^k$$ has a closed form which involves the hypergeometric function. Playing with this summation, I arrived to something "rather" simple which write $$S(m)=m (n+1) \left((-1)^n-\binom{2 n+1}{n+1} (m-1)^{n+2} m^n \, _2F_1(1,2n+2;n+1;m)\right)$$ I do not see how a simpler rigorous closed form could be obtained.


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