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Prove \(\binom{n}{0}^2 + \binom{n}{1}^2 + \cdots + \binom{n}{n}^2 = \binom{2n}{n}\)

Guest Jun 28, 2018
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I just used several values for n and all of them show that your expression is TRUE!!. But I don't know how to prove it! Sorry about that.

Guest Jun 29, 2018
edited by Guest  Jun 29, 2018

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