If $\displaystyle n\ge 0$, show that

$\displaystyle \displaystyle \sum_{k=0}^{n}\binom{2n}{k}\binom{2n-2k}{n-k} = \sum_{k=0}^{n}\binom{2n+1}{k} = \sum_{k=0}^{2n}\binom{2n}{k}$

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- Aug 11th 2010, 07:33 AMVandermondeAn Equality of Binomial SumsIf $\displaystyle n\ge 0$, show that

$\displaystyle \displaystyle \sum_{k=0}^{n}\binom{2n}{k}\binom{2n-2k}{n-k} = \sum_{k=0}^{n}\binom{2n+1}{k} = \sum_{k=0}^{2n}\binom{2n}{k}$ - Aug 13th 2010, 01:53 PMawkward
- Aug 16th 2010, 12:25 AMVlasev
The first one seems to the the odd one out. The last two evaluate to the same closed form expression. I suppose there is a typo in the first sum.