Prove that $(^nC_0)^2 + (^nC_1)^2 + (^nC_2)^2 + \cdots + (^nC_n)^2 = {^{2n}C}_n$

Asked Aug 22 '16 at 18:23

Active Aug 22 '16 at 18:30

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Can anyone help me prove that in the most simple way possible?

edited Aug 22 '16 at 18:30

Michael Hardy

asked Aug 22 '16 at 18:23

Youssef Walid

Not sure how many duplicates exist for this question. Here is one: http://math.stackexchange.com/q/148583/321264. – StubbornAtom Aug 22 '16 at 18:24

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