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Here's the question:

Prove that for all positive integers k ≤ n,

$\sum_{i=0}^{k} {n \choose i} (-1)^i = {n -1 \choose k}(-1)^k$

So far, I've noticed that you can change $ {n \choose i}$ into ${n - 1 \choose i} + {n -1 \choose i - 1}$, which gets me closer to where I want to be, but beyond that I'm stuck.

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    Are you familiar with telescoping series?2012-10-15
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    To some extent, but I don't believe we've seen them in the context of this class.2012-10-15
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    If you take the relation you noted, plug it in and write it out, you get a telescoping series in $(-1)^i\binom{n-1}{i}$.2012-10-15

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