1. Let $\binom{n}{j}=\frac{n!}{j!(n-j)!}$, then the value of $\sum_{k=0}^{49}(-1)^{k}\binom{99}{2 k}$ is
A. $-2^{50}$
B. $-2^{49}$
C. $2^{49}$
D. $2^{50}$
1. B Consider the expansion of $(1+\mathrm{i})^{99}$, on one hand, $(1+\mathrm{i})^{99}=\sum_{r=0}^{99} C_{99}^{r} \mathrm{i}^{r}$, its real part is $\sum_{k=0}^{49}(-1)^{k} C_{99}^{2 k}$, on the other hand, $(1+\mathrm{i})^{99}=(1+\mathrm{i})(1+\mathrm{i})^{98}=(1+\mathrm{i})(2 \mathrm{i})^{49}=-2^{49}+2^{49} \mathrm{i}$. By comparing the real parts, we can solve it.