Find the sum of $\binom{2007}{0}+\binom{2007}{4}+…+\binom{2007}{2004}$ [duplicate]

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

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

• I think there's a typo, how can 2007 be the last term? Shouldn't it be 2004 (a multiple of 4)? – shardulc says Reinstate Monica Apr 27 '16 at 7:02
• @shardulc: Thank you! – Roman83 Apr 27 '16 at 7:04
• 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? – learner Apr 27 '16 at 7:06
• I think this, this, and this may be useful. (Perhaps even this though I'm not sure.) – shardulc says Reinstate Monica Apr 27 '16 at 7:08
• @shardulc Yes. In other words this may be a duplicate question. – almagest Apr 27 '16 at 7:12