$$
10 \cdot 81 S=1-\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}-\frac{1}{32}+\frac{1}{64}-\frac{1}{128}-\cdots,
$$

Then $S$ equals
(A) 0.
(B) $\frac{2}{7}$.
(C) $\frac{6}{7}$.
(D) $\frac{9}{32}$.
(E) $\frac{27}{32}$.
(10th American High School Mathematics Examination, 1959)
[Solution 1] Transform the given expression as follows:
$$
\begin{aligned}
S & =\left(1-\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{8}-\frac{1}{16}-\frac{1}{32}\right)+\left(\frac{1}{64}-\frac{1}{128}-\frac{1}{256}\right)+\cdots \\
& =\frac{1}{4}+\frac{1}{32}+\frac{1}{256}+\cdots,
\end{aligned}
$$
$$
\therefore \quad S=\frac{\frac{1}{4}}{1-\frac{1}{8}}=\frac{2}{7} \text {. }
$$

Therefore, the answer is (B).
[Solution 2] Let $S_{1}=1+\frac{1}{8}+\frac{1}{64}+\cdots, S_{2}=\frac{1}{2}+\frac{1}{16}+\frac{1}{128}+\cdots$,
$$
S_{3}=\frac{1}{4}+\frac{1}{32}+\frac{1}{256}+\cdots \text {, }
$$

then
$$
\begin{array}{l}
S=S_{1}-S_{2}-S_{3}=\frac{1}{1-\frac{1}{8}}-\frac{\frac{1}{2}}{1-\frac{1}{8}}-\frac{\frac{1}{4}}{1-\frac{1}{8}} \\
=\frac{1-\frac{1}{2}-\frac{1}{4}}{1-\frac{1}{8}}=\frac{2}{7} .
\end{array}
$$

Therefore, the answer is $(B)$.