10.83 A circle is inscribed in a square with side length $m$, another square is inscribed in this circle, and then another circle is inscribed in the second square, and so on. If $S_{n}$ represents the sum of the areas of the first $n$ circles, as $n$ increases indefinitely, $S_{n}$ approaches
(A) $\frac{\pi m^{2}}{2}$.
(B) $\frac{3 \pi m^{2}}{8}$.
(C) $\frac{\pi m^{2}}{3}$.
(D) $\frac{\pi m^{2}}{4}$.
(E) $\frac{\pi m^{2}}{8}$.
(18th American High School Mathematics Examination, 1967)
[Solution] Let the side length of the $k$-th square be $a_{k}$. According to the problem, we have
$$
a_{1}=m, \quad a_{k}=\frac{\sqrt{2}}{2} a_{k-1}(k=2,3, \cdots) .
$$

Let the radius of the $k$-th circle be $r_{k}$. According to the problem, we also have
$$
\begin{array}{l}
r_{1}=\frac{1}{2} a_{1}=\frac{1}{2} m, \\
r_{k}=\frac{1}{2} a_{k}=\frac{1}{2}\left(\frac{\sqrt{2}}{2}\right)^{k-1} m .
\end{array}
$$

Then the area of the $k$-th circle is
$$
A_{k}=\pi r_{k}^{2}=\pi\left(\frac{1}{2}\right)^{k-1}\left(\frac{m}{2}\right)^{2},
$$

Thus, $S_{n}=\pi\left(\frac{m}{2}\right)^{2}+\pi \cdot \frac{1}{2}\left(\frac{m}{2}\right)^{2}+\pi \cdot \frac{1}{4}\left(\frac{m}{2}\right)^{2}+\cdots$
$$
\begin{aligned}
& +\pi \cdot \frac{1}{2^{n-1}}\left(\frac{m}{2}\right)^{2} \\
= & \frac{\pi m^{2}}{4}\left(1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^{n-1}}\right) \\
= & \frac{\pi m^{2}}{4} \cdot 2\left[1-\left(\frac{1}{2}\right)^{n}\right] .
\end{aligned}
$$

Therefore, $\lim _{n \rightarrow \infty} S_{n}=\frac{\pi m^{2}}{2}$.
Hence, the answer is $(A)$.