Example 8 The sequence $a_{n}$ is defined as follows: $a_{1}=0, a_{2}=1$
$$
a_{n}=\frac{1}{2} n a_{n-1}+\frac{1}{2} n(n-1) a_{n-2}+(-1)^{n}\left(1-\frac{n}{2}\right)(n \geqslant 3)
$$

Try to find the simplest expression for $f_{n}=a_{n}+2 C_{n}^{1} a_{n-1}+3 C_{n}^{n} a_{n-2}+\cdots+(n-1) C_{n}^{n-2} a_{2}+n C_{n}^{n-1} a_{1}$.
Solve the first few terms: $a_{3}=2, a_{4}=9, a_{5}=44$
Let the number of derangements of the first $n$ positive integers be $D_{n}$, then $D_{n}=n!\left[1-\frac{1}{1!}+\frac{1}{2!}+\cdots+(-1)^{n} \frac{1}{n!}\right]$. Notice that $a_{k}=D_{k}(k=1,2,3,4,5)$. We will prove that $a_{n}=D_{n}\left(n \in \mathbf{N}^{*}\right)$
Since $D_{n}-n D_{n-1}=(-1)^{n}$, we will prove by induction that $a_{n}-n a_{n-1}=(-1)^{n}$
Verification shows that $a_{2}-2 a_{1}=1=(-1)^{2}, a_{3}-3 a_{2}=-1=(-1)^{3}, a_{4}-4 a_{3}=1=(-1)^{\prime}$
Assume the proposition holds for $n$, i.e., $a_{n}-n a_{n-1}=(-1)^{n}$
Then
$$
\begin{aligned}
a_{n-1} & =\frac{n+1}{2} a_{n}+\frac{n(n+1)}{2} a_{n-1}+(-1)^{n-1}\left(1-\frac{n+1}{2}\right) \\
& =\frac{n+1}{2} a_{n}+\frac{(n+1)}{2}\left[a_{n}-(-1)^{n}\right]+(-1)^{n+1}\left(1-\frac{n+1}{2}\right) \\
& =(n+1) a_{n}+(-1)^{n+1}
\end{aligned}
$$

Thus, the proposition holds for $n+1$.
Consider the combinatorial meaning of $C_{n}^{*} a_{n-k}(k=0,1,2, \cdots, n-1)$, it represents the number of ways to arrange the first $n$ positive integers $1,2, \cdots, n$ such that exactly $k$ numbers $a_{1}{ }^{\prime}, a_{2}{ }^{\prime}, \cdots, a_{k}{ }^{\prime}$ are placed in the $a_{1}{ }^{\prime}, a_{2}{ }^{\prime}, \cdots, a_{k}{ }^{\prime}$ positions, respectively.

Therefore, $a_{n}+C_{n}^{1} a_{n-1}+C_{n}^{2} a_{n-2}+\cdots+C_{n}^{n-2} a_{2}+C_{n}^{n-1} a_{1}$ represents the number of all permutations of the positive integers $1,2, \cdots, n$ except for $(1,2, \cdots, n)$, which is $n!-1$. Hence,
$$
a_{n}+C_{n}^{1} a_{n-1}+C_{n}^{2} a_{n-2}+\cdots+C_{n}^{n-2} a_{2}+C_{n}^{n-1} a_{1}=n!-1
$$

Thus, we have
$$
\begin{array}{l}
C_{n}^{1} a_{n-1}+2 C_{n}^{2} a_{n-2}+\cdots+(n-1) C_{n}^{n-1} a_{1}=n\left(a_{n-1}+C_{n-1}^{1} a_{n-2}+\cdots+C_{n-1}^{n-1} a_{1}\right) \\
=n[(n-1)!-1]
\end{array}
$$
(1) + (2) gives $a_{n}+2 C_{n}^{1} a_{n-1}+3 C_{n}^{2} a_{n-2}+\cdots+(n-1) C_{n}^{n-2} a_{2}+{ }_{n} C_{n}^{n-1} a_{1}=2 n!-(n+1)$.
That is,
$$
f_{n}=2 n!-(n+1)
$$