$10 \cdot 79 \frac{1}{7}+\frac{2}{7^{2}}+\frac{1}{7^{3}}+\frac{2}{7^{4}}+\cdots$ to infinity has the sum
(A) $\frac{1}{5}$.
(B) $\frac{1}{24}$.
(C) $\frac{5}{48}$.
(D) $\frac{1}{16}$.
(E) None of the above answers is correct.
(1st American High School Mathematics Examination, 1950)
[Solution]
$$
\begin{aligned}
\text { Original expression } & =\left(\frac{1}{7}+\frac{1}{7^{3}}+\frac{1}{7^{5}}+\cdots\right)+\left(\frac{2}{7^{2}}+\frac{2}{7^{4}}+\frac{2}{7^{6}}+\cdots\right) \\
& =\frac{\frac{1}{7}}{1-\frac{1}{7^{2}}}+\frac{\frac{2}{7^{2}}}{1-\frac{1}{7^{2}}}=\frac{7}{48}+\frac{2}{48}=\frac{3}{16} .
\end{aligned}
$$

Therefore, the answer is $(E)$.