This sum popped out of one of my calculations, I know what it should evaluate to, but I have no idea how to prove it.
$$\sum_{i=0}^{r}{n \choose 2i } - {n\choose 2i -1}$$
I know that $2i-1$ is negative for $i=0$, but for the purpose of this sum, we will say that ${n \choose x}=0$ if $n<0$ or $x<0$. So this sum is basically summing the difference of consecutive even/odd binomial coefficient pairs. We can rewrite this sum as
$$\sum_{i=0}^{2r}(-1)^i {n \choose i}.$$
I don't really know how to proceed from here, I couldn't find any information on evaluating sums of alternating series involving binomial coefficients.
Answer
This is a special case of the result
$$\sum_{k=0}^m(-1)^k{n\choose k}=(-1)^m{n-1\choose m}$$
($0\le m\le n-1$) which can be proved by induction on $m$.
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