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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Originally Posted by Vandermonde If $\displaystyle n\ge 0$, show that $\displaystyle \displaystyle \sum_{k=0}^{n}\binom{2n}{k}\binom{2n-2k}{n-k} = \sum_{k=0}^{n}\binom{2n+1}{k} = \sum_{k=0}^{2n}\binom{2n}{k}$ This is false for n=2.
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.
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