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I got the result from wolfram alpha: $\sum\limits_{i=k}^n {{i-1}\choose{k-1}} = \frac{(n+1-k){{n} \choose {k-1}}}{k}$, but I can't see how it derived it. The expression could be slightly simplified: $\sum\limits_{i=k-1}^{n-1} {{i}\choose{k-1}} = \sum\limits_{i=m}^{n-1} {{i}\choose{m}}\ ;\ m=k-1$. But I still can't see a solution.

galra
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  • Remember the recurrence relation for binomial coefficients: ${{n+1}\choose {k+1}}={{n} \choose {k+1}} + {n \choose k}$. – Gabriel Burns Nov 14 '16 at 17:54
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    This looks like the same as http://math.stackexchange.com/questions/833451/prove-sum-i-0n-binomik-1k-1-binomnkk-a-k-a-hockey-stick-ident. – StubbornAtom Nov 14 '16 at 18:06

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