How do we show that this identity holds for any n? Any hints or solutions?
Show $\sum_{ r = 0}^{n} {(\binom n r)}^{2} = \frac{(2n)!}{(n!)^{2}}$
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Edited question. – islamfaisal Jun 23 '15 at 06:10
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A broad hint: write one of your two $n\choose r$ terms as $n\choose n-r$ using the symmetry of the binomial coefficients. From there you should be able to apply familiar identities... – Steven Stadnicki Jun 23 '15 at 06:16
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Already tried this but still stuck, can you give more hints? – islamfaisal Jun 23 '15 at 06:18
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See Vandermonde's identity. – Lucian Jun 23 '15 at 06:23
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For $~n=\dfrac12~$ we have $~\displaystyle\sum_{n=0}^\infty\frac{\displaystyle{2n\choose n}^2}{16^n~(2n-1)^2}=\frac4\pi$ – Lucian Jun 23 '15 at 06:28