Prove $\binom{2n}{n}^2 = \sum^{n}_{k = 0} \binom{2n}{2k} \binom{2k}{k} \binom{2n - 2k}{n - k}$

Asked Nov 06 '23 at 19:40

Active Nov 06 '23 at 19:51

Viewed 79 times

Actually I invented this formula. Can someone check if my proof is correct? Or can someone come up with a new proof? Thank you.

edited Nov 06 '23 at 19:51

Gonçalo

asked Nov 06 '23 at 19:40

Yuding Liu

1

Welcome to MSE. Please fully display your mathematical content on this post rather than with a link. – Axel Nov 06 '23 at 19:42
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Or https://math.stackexchange.com/q/4042869/694754 – Brian Moehring Nov 06 '23 at 19:48
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The formula is already known. Please first check the literature before posting. Naming a known identity after yourself is not right. – Dietrich Burde Nov 06 '23 at 19:59
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As a side-note: It's great that you discovered that identity for yourself, but I'd strongly advise against ever naming an identity/theorem/theory for yourself. In my experience, the broader mathematical community responds unfavorably to that. – Brian Moehring Nov 06 '23 at 19:59
Thank you all for your help. Thank you for pointing me to the existing questions. – Yuding Liu Nov 08 '23 at 02:59

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