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

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)?

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

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