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$ \sum_{k=r}^{n} {n \choose k} = \sum_{k = r - 1}^{n-1}{k \choose r -1}2^{n-1-k} $

The proof is trivial but I haven't seen this identity anywhere. Perhaps it's a special case of a more general identity?

aspy591
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  • http://math.stackexchange.com/questions/84746/bounds-on-sum-k-0m-binomnkxk-and-sum-k-0m-binomnkxk1 – Alex Apr 27 '13 at 15:56

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Whatever sums of hypergeometric terms are around can be solved by the techniques of Petkovsek, Wilf, Zeilberger's "A = B"

vonbrand
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