10-82 A line was initially 1 inch long and grew according to the following rule (where the first term is the initial length).
$$
1+\frac{1}{4} \sqrt{2}+\frac{1}{4}+\frac{1}{16} \sqrt{2}+\frac{1}{16}+\frac{1}{64} \sqrt{2}+\frac{1}{64}+\cdots \cdots
$$

If the growth process continues indefinitely, then the limit of the line length is
(A) $\infty$.
(B) $\frac{4}{3}$.
(C) $\frac{8}{3}$.
(D) $\frac{1}{3}(4+\sqrt{2})$.
(E) $\frac{2}{3}(4+\sqrt{2})$.
(3rd American High School Mathematics Examination, 1952)
[Solution]
$$
\begin{aligned}
\text { Original expression } & =1+\frac{1}{4}(\sqrt{2}+1)+\frac{1}{16}(\sqrt{2}+1)+\frac{1}{64}(\sqrt{2}+1)+\cdots \\
& =1+\frac{\frac{1}{4}(\sqrt{2}+1)}{1-\frac{1}{4}}=\frac{4+\sqrt{2}}{3} .
\end{aligned}
$$

Therefore, the answer is $(D)$.