Ask 31 The sequence $\left\{a_{n}\right\}$ is defined as follows:
$$
\begin{array}{r}
a_{1}=0, a_{2}=1, \cdots, a_{n}=\frac{1}{2} n a_{n-1}+\frac{1}{2} n(n-1) a_{n-2}+(-1)^{n}\left(1-\frac{n}{2}\right) \\
(n \geqslant 3) .
\end{array}
$$

Try to find the simplest expression for $f_{n}=a_{n}+2 \mathrm{C}_{n}^{1} a_{n-1}+3 \mathrm{C}_{n}^{2} a_{n-2}+\cdots+(n-1) \mathrm{C}_{n}^{n-2} a_{2}+n \mathrm{C}_{n}^{n-1} a_{1}$.
Solve: First, calculate the first $n$ terms $a_{3}=2, a_{4}=9, a_{5}=44$, and $a_{6}=265$. This matches the famous derangement numbers. We can also add $a_{0}=1$, for example, 231 and 312 are derangements, so $a_{3}=2$. $a_{n}$ represents the number of derangements of $1-n$, and we have
$$
a_{n}+\mathrm{C}_{n}^{1} a_{n-1}+\mathrm{C}_{n}^{2} a_{n-2}+\cdots+\mathrm{C}_{n}^{n-2} a_{2}+\mathrm{C}_{n}^{n-1} a_{1}=n!-1 .
$$

Remove (1) from $f_{n}$.
$$
\begin{aligned}
& \mathrm{C}_{n}^{1} a_{n-1}+2 \mathrm{C}_{n}^{2} a_{n-2}+\cdots+(n-1) \mathrm{C}_{n}^{n-1} a_{1} \\
= & n\left[a_{n-1}+\mathrm{C}_{n-1}^{1} a_{n-2}+\cdots+\mathrm{C}_{n-1}^{n-2} a_{1}\right]=n((n-1)!-1) .
\end{aligned}
$$
Adding (1) and (2) gives
$$
n!-1+n((n-1)!-1)=2 n!-(n+1) \text {. }
$$

Now, we prove that the original sequence is indeed the derangement sequence. The derangement sequence satisfies the equation
$$
a_{n}=(n-1)\left(a_{n-1}+a_{n-2}\right) \text {. }
$$

This can be rewritten as
$$
\begin{aligned}
a_{n}-n a_{n-1} & =-\left(a_{n-1}-(n-1) a_{n-2}\right)=a_{n-2}-(n-2) a_{n-3} \\
& =\cdots=(-1)^{n-1}\left(a_{1}-a_{0}\right)=(-1)^{n} .
\end{aligned}
$$