# binomial coefficient

Prove that $$\frac{1}{\sqrt{1-4t}} \left(\frac{1-\sqrt{1-4t}}{2t}\right)^k = \sum\limits_{n=0}^{\infty}\binom{2n+k}{n}t^n, \quad \forall k\in\mathbb{N}.$$ I tried already by induction over $k$ but i have problems showing the statement holds for $k=0$ or $k=1$.

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See Concrete Mathematics, page 203. –  wj32 Nov 15 '12 at 8:34
I am also looking for a combinatorical proof of this identity. –  bronko Dec 7 '12 at 3:45
Thank you for your answer. I expanded the left hand side via binomial formula. The problem is that the sum on the left hand side only counts to k while the one on the right hand side counts to $\infty$. –  bronko Nov 22 '12 at 13:47
I expanded the left hand side to $\sum_{n=0}^k\binom{k}{n}\left(\frac{1}{2t}\right)^k\frac{-\sqrt{1-4t})^{k-n-1}}‌​{2t}^{k-n}}$ but that's it. I do not know how to proceed. –  bronko Nov 24 '12 at 8:06
$$\sum_{n=0}^k \binom{k}{n} \left( \frac{1}{2t} \right)^k (- \sqrt{1-4t})^{k-n-1} (2t)^{-k+n}$$ –  bronko Nov 26 '12 at 5:11