$10 \cdot 60$ algebraic expression
$$
\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\cdots+\frac{1}{(2 n-1)(2 n+1)}+\cdots+\frac{1}{255 \cdot 257}
$$

The value is
(A) $\frac{127}{255}$.
(B) $\frac{128}{255}$.
(C) $\frac{1}{2}$.
(D) $\frac{128}{257}$.
(E) $\frac{129}{257}$.
(28th American High School Mathematics Examination, 1977)
[Solution 1] Since $\frac{1}{k(k+2)}=\frac{1}{2}\left(\frac{1}{k}-\frac{1}{k+2}\right)$,
then
$$
\begin{array}{l}
\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\cdots+\frac{1}{(2 n-1)(2 n+1)} \\
=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\cdots+\frac{1}{2 n-1}-\frac{1}{2 n+1}\right) \\
=\frac{1}{2}\left(1-\frac{1}{2 n+1}\right)=\frac{n}{2 n+1},
\end{array}
$$

When $n=128$, the required sum is $\frac{128}{2 \cdot 128+1}=\frac{128}{257}$.
Therefore, the answer is $(D)$.
[Solution 2] $\because \frac{1}{n(n-k)}+\frac{1}{n(n+k)}=\frac{2}{(n-k)(n+k)}$, the required sum is
$$
\begin{aligned}
& 2\left(\frac{1}{1 \cdot 5}+\frac{1}{5 \cdot 9}+\cdots+\frac{1}{253 \cdot 257}\right) \\
= & 4\left(\frac{1}{1 \cdot 9}+\frac{1}{9 \cdot 17}+\cdots+\frac{1}{249 \cdot 257}\right) \\
= & \cdots=128\left(\frac{1}{1 \cdot 257}\right) .
\end{aligned}
$$

Therefore, the answer is $(D)$.