$34 \cdot 2$ 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) 0 .
(D) $2^{49}$.
(E) $2^{50}$.
(40th American High School Mathematics Examination, 1989)
[Solution] Notice that $(1+i)^{99}=\sum_{r=0}^{99} 1^{99-r} \cdot i^{r} \cdot\binom{99}{r}$, and
$$
i^{2}=-1, \quad i^{3}=-i, \quad i^{4}=1,
$$

then the real part of its expansion is $\sum_{k=0}^{49} i^{2 k}\binom{99}{2 k}=\sum_{k=0}^{49}(-1)^{k}\binom{99}{2 k}$.
And
$$
\begin{aligned}
(1+i)^{99} & =(1+i) \cdot(1+i)^{98}=(1+i)\left[(1+i)^{2}\right]^{49} \\
& =(1+i)(2 i)^{49}=2^{49} \cdot i^{49}(1+i) \\
& =2^{49}(i-1)=-2^{49}+i \cdot 2^{49},
\end{aligned}
$$

therefore
$$
\sum_{k=0}^{49}(-1)^{k}\binom{99}{2 k}=-2^{49} \text {. }
$$

Hence the answer is $(B)$.