$\begin{array}{c}\text { Example 2. Find the sum } S_{n}=\frac{1}{1 \cdot 5}+\frac{1}{5 \cdot 9}+\cdots \\ +\frac{1}{(4 n-3)(4 n+1)} .\end{array}$
$$
\text{Solve } \begin{aligned}
& S_{n}=\frac{1}{4}\left[\left(1-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{9}\right)\right. \\
& \left.+\cdots+\left(\frac{1}{4 n-3}-\frac{1}{4 n+1}\right)\right] \\
& =\frac{n}{4 n+1} .
\end{aligned}
$$

In general, if $a_{1}, a_{2}, \cdots, a_{n}, \cdots$ is an arithmetic sequence, $a_{k} \neq 0, k=1,2, \cdots$, with a common difference of $d$, then it is easy to obtain:
$$
\begin{array}{l}
\frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\cdots+\frac{1}{a_{n} a_{n+1}} \\
=\frac{n}{a_{1} a_{n+1}} \\
\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\cdots \\
+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}}=\frac{\sqrt{a_{n}}-\sqrt{a_{1}}}{d}
\end{array}
$$