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I would like to know if there is a closed form of $$\sum_{k=0}^{n}\frac{4^{k}}{\left(2k\right)!\left(n-k\right)!^{2}}.$$ Wolfram gives a strange closed form and, i.e., $$\frac{16^{n}\left(2n-\frac{1}{2}\right)!}{\sqrt{\pi}\left(2n!\right)^{2}}$$ and I'm not sure is right. If it's so, I have no idea how to prove it. Is it right? And how to prove this identity? Any help would be appreciate! Thank you.

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    $\begingroup$ You ask whether there is a closed form, and then you write that Wolfram gives a closed form, thereby answering your own question. Next! $\endgroup$ – Gerry Myerson May 31 '15 at 10:10
  • $\begingroup$ @GerryMyerson I would know hot to prove it! And what Wolfram wrote seems strange to me. $\endgroup$ – User May 31 '15 at 10:11
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    $\begingroup$ Then why not share the big secret of what Wolfram gave you? $\endgroup$ – Gerry Myerson May 31 '15 at 10:12
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    $\begingroup$ @GerryMyerson I really don't understand these ironic comments. However, I'll write what Wolfram gives. $\endgroup$ – User May 31 '15 at 10:14
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    $\begingroup$ @FundThmCalculus no, Gerry was referring to the initial edit. Elajan asked if there was a closed form and said that Wolfram gave one, but didn't say what Wolfram claimed it was until Gerry direction asked it. $\endgroup$ – davidlowryduda May 31 '15 at 10:23
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It seems like you, too, have come across Jack's question from yesterday. Assuming the validty of his identity, your sum, which is the convolution (Cauchy product) formed when multiplying $\cos(1)$ and $\sum_{k\geq 0}\frac{(-1)^k}{k!^2 4^k}$ is equal to $$\frac{(4n)!}{(2n)!^{3}}$$

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  • $\begingroup$ I hadn't seen that question. So is it true that $$\frac{16^{n}\left(2n-\frac{1}{2}\right)!}{\sqrt{\pi}\left(2n\right)!^{2}} = \frac {\left(4n\right)!}{\left(2n\right)!^{3}}?$$ $\endgroup$ – User May 31 '15 at 13:30
  • $\begingroup$ Indeed, put $z=2n+\frac{1}{2}$ in the Gamma function duplication formula $\endgroup$ – nospoon May 31 '15 at 15:03

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