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Show that $$\sum_{k=0}^n(-1)^k\binom nk=0$$ So for odd $n$ we have an even number of terms. So $\binom nk=\binom n{n-k}$ which have opposite signs. Thus the sum is 0.

For even $n$ we have that $$\sum_{k=0}^n(-1)^k\binom nk= \binom n0+\sum_{k=1}^{n-1}(-1)^k\binom nk+\binom nn$$ Now $$\sum_{k=1}^{n-1}(-1)^k\binom nk= \sum_{k=1}^{n-1}(-1)^k\left[\binom{n-1}k+\binom{n-1}{k-1}\right]$$ What would that sum be in the square brackets?

  • 1
    Write out that last sum for some small values of $n$. I think that should make things clearer.2011-12-27
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    A generalization: http://math.stackexchange.com/questions/4175/beautiful-identity-sum-k-mn-1k-m-binomkm-binomnk-delta2011-12-27
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    Related: [How do I count the subsets of a set whose number of elements is divisible by 3? 4?](http://math.stackexchange.com/q/918/) and [Counting subsets with r mod 5 elements](http://math.stackexchange.com/q/1382/)2014-01-15
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    It might be worth mentioning that for $n=0$ the sum is equal to one, not zero.2015-10-28

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