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.

  • $\begingroup$ Isn't this a telescoping sum that collapses down to just $a_{n/4}-a_0$, plus a couple other terms depending on the deivisibility of $n$ by $4$? $\endgroup$
    – Ovi
    Feb 28, 2018 at 7:37

2 Answers 2


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 $m$.


Consider the coefficient of $x^{m}$ in the expansion of $$(1-x)^n\frac{1}{1-x}=\sum_{r=0}^{n}(-1)^r{n \choose r}x^r\sum_{r=0}^{∞}x^r$$


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