Example 6 The sum of the first $n$ terms of the sequence $\left\{\frac{1}{n(n+1)(n+2)}\right\}$ is $\qquad$
Since $\frac{1}{n(n+2)}$ can be decomposed into the difference of two terms: $\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)$, we have
$$
\frac{1}{n(n+1)(n+2)}=\frac{1}{n(n+2)} \cdot \frac{1}{n+1}=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right) \cdot \frac{1}{n+1}
$$

Thus,
$$
2 \cdot \frac{1}{n(n+1)(n+2)}=\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}
$$

Let the sum of the first $n$ terms of the sequence be $S_{n}$, then
$$
\begin{aligned}
2 S_{n}= & \left(\frac{1}{1 \cdot 2}-\frac{1}{2 \cdot 3}\right)+\left(\frac{1}{2 \cdot 3}-\frac{1}{3 \cdot 4}\right)+\left(\frac{1}{3 \cdot 4}-\frac{1}{4 \cdot 5}\right)+ \\
& \cdots+\left[\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right] \\
= & \frac{1}{1 \cdot 2}-\frac{1}{(n+1)(n+2)}=\frac{n^{2}+3 n}{2(n+1)(n+2)}
\end{aligned}
$$

Therefore, $S_{n}=\frac{n^{2}+3 n}{4(n+1)(n+2)}$