$32 \cdot 8$ When $n$ is a positive integer, the sum of the decimal digits of $\left(10^{4 n^{2}+8}+1\right)^{2}$ is
(A) 4.
(B) $4 n$.
(C) $2+2 n$.
(D) $4 n^{2}$.
(E) $n^{2}+n+2$
(26th American High School Mathematics Examination, 1975)
[Solution] For any positive integer $k$, $10^{k}+1=100 \cdots 01$, and
$$
\begin{array}{r}
100 \cdots 01 \\
\times \quad 100 \cdots 01 \\
\hline 100 \cdots 01 \\
+\quad 100 \cdots 01 \\
\hline 100 \cdots 0200 \cdots 01
\end{array}
$$
$\therefore$ The sum of the digits is $1+2+1=4$. Therefore, the answer is $(A)$.