II. The sequence $\left\{a_{n}\right\}$ is defined as follows: $a_{1}=\hat{j}, a_{2}=1, a_{n}=$ $\frac{1}{2} n a_{n-1}+\frac{1}{2} n(n-1) a_{n-2}+(-1)^{n}\left(1-\frac{n}{2}\right), n \geqslant$ 3. Try to find the simplest expression for $i_{n}=a_{n}+2 C_{n}^{1} a_{n-1}+3 C_{n}^{2} a_{n-2}+\cdots+(n-1)$. $\mathrm{C}_{n}^{2-2} a_{2}+n \mathrm{C}_{n}^{-1} a_{1}$. (Supplied by Huang Yumin)
Solution 1: Let $b_{n}=\frac{a_{n}}{n!}$, then
$b_{1}=0, b_{2}=\frac{1}{2}, b_{3}=\frac{1}{3}$,
$b_{n}=\frac{1}{2}\left(b_{n-1}+b_{n-2}\right)+(-1) \cdot \frac{1-\frac{n}{2}}{n!}$
Let $g_{n}=\frac{1}{n} f_{n}=\frac{1}{n} \sum_{k=1}^{n}(n-k+1) G_{4}^{-a_{t}}$
$=\sum_{k=1}^{n} \frac{n-k+1}{(n-k)!} \cdot b_{k}$.
$\therefore g_{n+1}-g_{n}$
$=\sum_{k=1}^{n+1}\left(\frac{n-k+2}{(n+1-k)!} \cdot b_{k}-\sum_{k=1}^{n}\left(\frac{n-k+1}{(n-k)!} \cdot b_{t}\right)\right.$
$=\sum_{k=1}^{n+1} \frac{n-k+2}{(n+1-k)!} \cdot b_{k}-\sum_{k=1}^{n+1} \frac{n-k+2}{(n-k+1)!}$
$=\sum_{k=1}^{n+1} \frac{n-k+2}{(n+1-k)!}\left(b_{k}-b_{k-1}\right)$.
But let $d_{n}=(-2)^{n}\left(b_{n}-b_{n-1}\right)$, then
$d_{n}=d_{n-1}+\left(1-\frac{n}{2}\right) \frac{1}{n!}$
So $d_{2}=2$,
$d_{n}=2+\sum_{i=1}^{\infty}\left(1-\frac{l}{2}\right) \frac{l^{2}}{2!}=\frac{2 n}{n!}$
Thus $b_{n}-b_{n-1}=\frac{d_{n}}{(-2)^{n}}=\frac{2^{n}}{(-2)^{n} n!}=\frac{(-1)^{n}}{n!}$
$\therefore g_{n+1}-g_{n}=\sum_{k=1}^{n+1}\left(\frac{n+2-k}{(n+1-k)} \cdot \frac{(-1)^{k}}{k!}\right)$
$=\sum_{k=1}^{n}\left(\frac{(-1)^{k}}{(n-k)!k!}\right)+\sum_{k=1}^{n+1} \frac{(-1)^{k}}{(n+1-k)!k!}$
$=\frac{1}{n!} \sum_{k=1}^{n}(-1)^{k}\binom{n}{k}+\frac{1}{(n+1)!}$
$\cdot \sum_{k=1}^{n+1}(-1)^{k} \cdot\binom{n+1}{k}$
$\sum_{k=0}^{\infty}(-1)^{k}\binom{n}{k}=0$,
$\sum_{k=0}^{n+1}(-1)^{k}\binom{n+1}{k}=0$,
$\therefore g_{n+1}-g_{n}$
$=-\frac{1}{n!}(1-n)-\frac{1}{(n+1)!}(1-(n+1))$
$=\frac{1}{(n-1)!}-\frac{1}{(n+1)!}$
Since $g_{3}=2 b_{2}+b_{3}=\frac{4}{3}$,
$f_{n}=n!g_{n}$
$=n!\left(\sum_{k=1}^{\infty} \frac{1}{(k-2)!}-\sum_{k=1}^{n} \frac{1}{n} z_{3}\right)$
$=n!\left(s+\frac{1}{2!}+\frac{1}{3!}-\frac{x+1}{n!}\right)$
$=2 \cdot n!-(n+1)$.
Solution 2: Let $b_{n}=\frac{a_{n}}{n!}$ then $b_{1}=0, b_{2}=\frac{1}{2}$,
$b_{n}=\frac{1}{2} b_{n-1}+\frac{1}{2} b_{n-2}+(-1)^{n} \frac{1}{n!}$
$+\frac{1}{2}(-1)^{n-1} \frac{1}{(n-1)!}$
$=\frac{1}{2}\left(a_{n-1}+(-1)^{n} \frac{1}{n!}\right)+\frac{1}{2}\left(a_{n-2}+(-1)^{n} \frac{1}{n!}+(-1)^{n-1}\left(\frac{1}{(n-1)!}\right)\right)$.
$c_{n}=c_{n-2}+(-1)^{n} \frac{1}{n!}+(-1)^{n-1} \frac{1}{(n-1)!}$,
$n \geqslant 3$.
$\therefore c_{n}=\frac{1}{2}\left(c_{n-1}+(-1)^{n} \frac{1}{n!}\right)$
$+\frac{1}{2}\left(c_{n-2}+(-1)^{n} \frac{1}{n!}+(-1)^{n-1} \frac{1}{(n-1)!}\right)$
$n \geqslant 3$.
By uniqueness,
$b_{n}=c_{n}, n=1,2, \cdots$
Thus, $b_{n}=b_{n-1}+(-1)^{n} \frac{1}{n!}, n \geqslant 2$.
(1)

Let $\sigma_{n}=\frac{1}{n!} f_{n}$, then
$\sigma_{n}=b_{n}+2 b_{n-1}+\frac{3}{2!} b_{n-2}+\cdots+\frac{n-1}{(n-2)!} b_{2}$.
From the recurrence formula (1) we get
$\sigma_{n+1}=\sigma_{n}+S_{n}$,
where $S_{n}=\frac{n}{2!(n-1)!}-\frac{n-1}{3!(n-2)!}+\cdots$
$+(-1)^{n-1} \frac{3}{(n-1)!2!}+(-1)^{n} \frac{2}{n!}$
$+(-1)^{n+1} \frac{1}{(n+1)!}$.

Clearly,
```
(-1)
+(-1)^{n-1} n
(1-x)
=1}-\mp@subsup{\textrm{C}}{n+1}{1}x+\mp@subsup{\textrm{C}}{n+1}{2}x^{2}+
+(-1)^{n-1}\mp@subsup{\textrm{C}}{n+1}{n-1}x^{n-1}+(-1)^{n}\mp@subsup{\textrm{C}}{n+1}{n}x
1-C C
- C C
+(-1)^{n-1}\mp@subsup{\textrm{C}}{n}{n-1}+(-1)^{n}\mp@subsup{\textrm{C}}{n}{n}]
=0.
Thus,
(-1)^{n+1}(n+1)!S_{n}
This gives
S_{n}=\frac{n^2+n-1}{(n+1)!}=\frac{n+1}{n!}-\frac{n+2}{(n+1)!}
Let d_{n}=\sigma_{n}+\frac{n+1}{n!}, then d_{n+1}=d_{n}. Thus,
d_{n}=d_{3}=\sigma_{3}+\frac{4}{3!}=a_{3}+2a_{2}+\frac{2}{3}=2
.\sigma
```