$10 \cdot 71$ product $\left(1-\frac{1}{2^{2}}\right)\left(1-\frac{1}{3^{2}}\right) \cdots\left(1-\frac{1}{9^{2}}\right)\left(1-\frac{1}{10^{2}}\right)$ equals
(A) $\frac{5}{12}$.
(B) $\frac{1}{2}$.
(C) $\frac{11}{20}$.
(D) $\frac{2}{3}$.
(E) $\frac{7}{10}$.
(37th American High School Mathematics Examination, 1986)
［Solution］Notice the following transformation:
$$
\begin{aligned}
\text { Original expression= } & \left(1-\frac{1}{2}\right)\left(1+\frac{1}{2}\right) \cdot\left(1-\frac{1}{3}\right)\left(1+\frac{1}{3}\right) \cdots \cdot\left(1-\frac{1}{10}\right)(1+ \\
& \left.\frac{1}{10}\right) \\
= & \frac{1}{2} \cdot \frac{3}{2} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{3}{4} \cdot \frac{5}{4} \cdots \cdot \frac{8}{9} \cdot \frac{10}{9} \cdot \frac{9}{10} \cdot \frac{11}{10} \\
= & \frac{11}{20} . \quad \text { (Note the cancellation of terms) }
\end{aligned}
$$

Therefore, the answer is $(C)$.