Example 4. In the sequence $\left\{a_{\mathrm{n}}\right\}$, $a_{2}=1, a_{3}=2$, and it satisfies: $a_{\mathrm{r}}=(n-1)\left(a_{\mathrm{n}-1}+a_{n-2}\right)(a n \geqslant 4)$, try to find the general term formula.
Let $a_{\mathrm{n}}=n!b_{\mathrm{n}}(n \geqslant 4)$, then we have $n!b_{\mathrm{n}}=(n-1)\left[(n-1)!b_{\mathrm{n}-1}+\right.$ $(n-2)!\cdot b_{n-2}$. Therefore, $n b_{n}=(n-1) b_{n-1} + b_{n-2}$. We get $b_{n}-b_{n-1}=-\frac{1}{n}\left(b_{n-1}-b_{n-2}\right)$ $(n \geq 4)$. By recursion, we have:
$$
\begin{array}{l}
b_{4}-b_{3}=-\frac{1}{4}\left(b_{3}-b_{2}\right), \\
b_{5}-b_{4}=-\frac{1}{5}\left(b_{4}-b_{3}\right), \\
\cdots \cdots \\
b_{n}-b_{n-1}=-\frac{1}{n}\left(b_{n-1}-b_{n-2}\right) .
\end{array}
$$

Multiplying the above equations and canceling out the intermediate terms, we get:
$$
\begin{array}{l}
b_{\mathrm{n}}-b_{\mathrm{n}-1}=\left(-\frac{1}{4}\right) \cdot\left(-\frac{1}{5}-\ldots . .\right. \\
\cdot\left(-\frac{1}{n}\right) \cdot\left(b_{\mathrm{3}}-b_{2}\right) .
\end{array}
$$

But
$$
\begin{aligned}
\dot{b}_{3}=\frac{a_{3}}{3!} & =\frac{1}{3}, \quad b_{2}=\frac{a_{2}}{2!}=\frac{1}{2}, \\
\therefore b_{3}-b_{2} & =\frac{1}{3}-\frac{1}{2}=-\frac{1}{6} \\
& =(-1)\left(-\frac{1}{2}\right)\left(-\frac{1}{3}\right), \\
b_{n}-b_{n-1} & =(-1)^{2} \cdot \frac{1}{n!} .(n \geqslant 4)
\end{aligned}
$$

Using the recursive summation method, we get:
$$
\begin{array}{c}
\sum_{\mathrm{k}=1}^{\mathrm{n}}\left(b_{\mathrm{k}}-b_{\mathrm{k}-1}\right)=b_{\mathrm{n}}-b_{3} \\
=\sum_{\mathrm{k}=4}^{\mathrm{n}} \frac{(-1)^{k}}{k!} . \\
\text { And } b_{3}=\frac{1}{6}=\frac{(-1)^{2}}{2!}+\frac{(-1)^{3}}{3!} . \\
\text { Therefore, } b_{\mathrm{n}}=\sum_{\mathrm{k}=2}^{\mathrm{n}} \frac{(-1)^{\mathrm{k}}}{k!}, \\
\text { i.e., } a_{\mathrm{n}}=n!\sum_{\mathrm{k}=2}^{\mathrm{n}} \frac{(-1)^{\mathrm{k}}}{k!} .
\end{array}
$$