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- Jan 26, 2012

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**Problem**: Prove that for $n>0$, \(\displaystyle \sum_{i=0}^{n} (-1)^i \binom{n}{i}=0\)

**Attempt**: This seems clearly like a proof based on induction.

1) Base case: for $n=1$, \(\displaystyle \sum_{i=0}^{1}(-1)^i \binom{1}{i}=(-1)^0 \binom{1}{0}+(-1)^1 \binom{1}{1}=1-1=0\)

2) Show that $n=k$ being valid implies $n=k+1$ is valid. Assume that \(\displaystyle \sum_{i=0}^{k} (-1)^i \binom{k}{i}=0\). Now I need to show that \(\displaystyle \sum_{i=0}^{k+1} (-1)^i \binom{k+1}{i}=0\)?

Am I right so far? The hint I'm given is to use the binomial theorem but there is something I'm missing about showing this is true. Can someone give me a small push (not the full proof please)?