Example 3 Given
$$
a_{k}=\frac{k+2}{k!+(k+1)!+(k+2)!} \text {. }
$$

Then the sum of the first 100 terms of the sequence $\left\{a_{n}\right\}$ is $\qquad$
(2006, Shanghai Jiao Tong University Independent Admission Examination)
Notice that,
$$
\begin{aligned}
a_{k} & =\frac{k+2}{k!+(k+1)!+(k+2)!} \\
& =\frac{k+2}{k![1+(k+1)+(k+2)(k+1)]} . \\
& =\frac{k+2}{k!(k+2)^{2}}=\frac{1}{k!(k+2)} \\
& =\frac{k+1}{(k+2)!}=\frac{1}{(k+1)!}-\frac{1}{(k+2)!} .
\end{aligned}
$$

Therefore, $\sum_{k=1}^{100} a_{k}=\frac{1}{2}-\frac{1}{102!}$.
[Note] Related to permutations and combinations, there are also split terms such as
$$
n \cdot n!=(n+1)!-n!, \mathrm{C}_{n}^{k}=\mathrm{C}_{n+1}^{k+1}-\mathrm{C}_{n}^{k+1}
$$

and so on.