(2) The sequence $\left\{a_{n}\right\}$ 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 $f_{n}=a_{n}+$ $2 \mathrm{C}_{n}^{1} a_{n-1}+3 \mathrm{C}_{n}^{2} a_{n-2}+\cdots+(n-1) \mathrm{C}_{n}^{n-2} a_{2}+n \mathrm{C}_{n}^{n-1} a_{1}$.
Let $b_{n}=\frac{a_{n}}{n!}$, then
$$
\begin{array}{c}
b_{1}=0, b_{2}=\frac{1}{2}, b_{3}=\frac{1}{3} \\
b_{n}=\frac{1}{2}\left(b_{n-1}+b_{n-2}\right)+(-1)^{n} \frac{1-\frac{n}{2}}{n!}
\end{array}
$$

Let
$$
\begin{aligned}
g_{n}= & \frac{1}{n!} f_{n}=\frac{1}{n!} \sum_{k=1}^{n}(n-k+1) \mathrm{C}_{n}^{n-k} a_{k}= \\
& \sum_{k=1}^{n} \frac{n-k+1}{(n-k)!} b_{k}
\end{aligned}
$$

So
$$
\begin{array}{l}
g_{n+1}-g_{n}= \\
\sum_{k=1}^{n+1} \frac{n-k+2}{(n+1-k)!} b_{k}-\sum_{k=1}^{n} \frac{n-k+1}{(n-k)!} b_{k}= \\
\sum_{k=2}^{n+1} \frac{n-k+2}{(n+1-k)!} b_{k}-\sum_{k=2}^{n+1} \frac{n-k+2}{(n-k+1)!} b_{k-1}= \\
\sum_{k=2}^{n+1} \frac{n-k+2}{(n+1-k)!}\left(b_{k}-b_{k-1}\right)
\end{array}
$$

But let $d_{n}=(-2)^{n}\left(b_{n}-b_{n-1}\right)$, then
$$
d_{n}=d_{n-1}+\left(1-\frac{n}{2}\right) \frac{2^{n}}{n!}
$$

So
$$
\begin{array}{c}
d_{2}=2 \\
d_{n}=2+\sum_{l=3}^{n}\left(1-\frac{l}{2}\right) \frac{2^{l}}{l!}=\frac{2^{n}}{n!}
\end{array}
$$

Hence
$$
b_{n}-b_{n-1}=\frac{d_{n}}{(-2)^{n}}=\frac{2^{n}}{(-2)^{n} n!}=\frac{(-1)^{n}}{n!}
$$

So
$$
\begin{array}{l}
g_{n+1}-g_{n}=\sum_{k=2}^{n+1} \frac{n+2-k}{(n+1-k)!} \cdot \frac{(-1)^{k}}{k!}= \\
\sum_{k=2}^{n} \frac{(-1)^{k}}{(n-k)!k!}+\sum_{k=2}^{n+1} \frac{(-1)^{k}}{(n+1-k)!k!}= \\
\frac{1}{n!} \sum_{k=2}^{n}(-1)^{k}\binom{n}{k}+\frac{1}{(n+1)!} \text {. } \\
\sum_{k=2}^{n+1}(-1)^{k} \cdot\binom{n+1}{k} \\
\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}=0 \\
\sum_{k=0}^{n+1}(-1)^{k}\binom{n+1}{k}=0 \\
\end{array}
$$

Because

So
$$
\begin{aligned}
g_{n+1}-g_{n}= & -\frac{1}{n!}(1-n)-\frac{1}{(n+1)!}(1-(n+1))= \\
& \frac{1}{(n-1)!}-\frac{1}{(n+1)!}
\end{aligned}
$$

Since $g_{3}=2 b_{2}+b_{3}=\frac{4}{3}$, then
$$
\begin{aligned}
f_{n}= & n!g_{n}=n!\left(\sum_{k=4}^{n} \frac{1}{(k-2)!}-\sum_{k=4}^{n} \frac{1}{k!}+g_{3}\right)= \\
& n!\left(g_{3}+\frac{1}{2!}+\frac{1}{3!}-\frac{n+1}{n!}\right)= \\
& 2 \cdot n!-(n+1)
\end{aligned}
$$