34.5 For any real number $a$ and positive integer $k$, define
$\left[\begin{array}{l}a \\ k\end{array}\right]=a(a-$ then $\left[\begin{array}{c}-\frac{1}{2} \\ 100\end{array}\right] \div\left[\begin{array}{c}\frac{1}{2} \\ 100\end{array}\right]$ equals
(A) -199 .
(B) -197 .
(C) -1 .
(D) 197 .
(E) 199 .
(39th American High School Mathematics Examination, 1988)
［Solution］According to the problem, we have
$$
\begin{aligned}
\text { Original expression } & =\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right) \cdots\left[-\frac{1}{2}-(100-1)\right] \frac{1}{100!}}{\left(\frac{1}{2}\right)\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right) \cdots\left[\frac{1}{2}-(100-1)\right] \frac{1}{100!}} \\
& =\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right) \cdots\left(-\frac{1}{2}-98\right)\left(-\frac{1}{2}-99\right)}{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) \cdots\left(\frac{1}{2}-98\right)\left(\frac{1}{2}-99\right)} \\
& =\frac{-\frac{1}{2}-99}{\frac{1}{2}}=-199 .
\end{aligned}
$$

Therefore, the answer is $(A)$.