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Find $$\binom{999}{0}-\binom{999}{2}+\binom{999}{4}-\binom{999}{6}+\cdots +\binom{999}{996}-\binom{999}{998}$$

A.$-2^{500}$ B.$-2^{499}$ C.$2^{500}$ D.$2^{499}$

By the way, I want to ask is there any command to type the combination sign(C)directly with (n,k)?

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    $\begingroup$ $\binom{999}{2}$ as an example to your second question (dollar, backslash, "binom{900}{2}", dollar). $\endgroup$ – barak manos Nov 20 '15 at 9:55
  • $\begingroup$ C^{n}_{k} produces $C^{n}_{k}$ while { n \choose k} and \binom{n}{k} produce ${ n \choose k}$ and $\binom{n}{k}$ $\endgroup$ – Henry Nov 20 '15 at 11:37
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HINT :

We have $$(1+i)^n=\binom n0+i\binom n1-\binom n2-i\binom n3+\binom n4+\cdots\tag1$$

Note here that $\binom n0-\binom n2+\binom n4-\cdots$ is the real part of $(1)$.

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    $\begingroup$ And consider that $(1+i)^2=2 i$. $\endgroup$ – DanielWainfleet Nov 20 '15 at 11:03

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