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This question already has an answer here:

Find the sum of $$S=\binom{2007}{0}+\binom{2007}{4}+\binom{2007}{8}+...+\binom{2007}{2004}$$

My work so far: $$(1+1)^n=2^n=\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+...+\binom{n}{n}$$ $$(1-1)^n=0=\binom{n}{0}-\binom{n}{1}+\binom{n}{2}-...+(-1)^n\binom{n}{n}$$

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marked as duplicate by almagest, Community♦ Apr 27 '16 at 7:13

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    $\begingroup$ I think there's a typo, how can 2007 be the last term? Shouldn't it be 2004 (a multiple of 4)? $\endgroup$ – shardulc Apr 27 '16 at 7:02
  • $\begingroup$ @shardulc: Thank you! $\endgroup$ – Roman83 Apr 27 '16 at 7:04
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    $\begingroup$ Defining $f(x)=(1+x)^{2007}$, your sum $\sum\limits_{i=0}^{501}\dbinom n{4i}$ equals $\dfrac 14\bigg(f(1)+f(-1)+f(i)+f(-i)\bigg)$. Do you see why? $\endgroup$ – learner Apr 27 '16 at 7:06
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    $\begingroup$ I think this, this, and this may be useful. (Perhaps even this though I'm not sure.) $\endgroup$ – shardulc Apr 27 '16 at 7:08
  • $\begingroup$ @shardulc Yes. In other words this may be a duplicate question. $\endgroup$ – almagest Apr 27 '16 at 7:12