数列 $\{a_n\}$ 定义如下：$a_{1}=0, a_{2}=1, a_{n}=\frac{1}{2} n a_{n-1}+\frac{1}{2} n(n-1) a_{n-2}+(-1)^{n}\left(1-\frac{n}{2}\right), n \geqslant 3$．试求 $f_{n}=a_{n}+2 C_{n}^{1} a_{n-1}+3 C_{n}^{2} a_{n-2}+\cdots+(n-1) C_{n}^{a-2} a_{2}+n C_{n}^{n-1} a_{1}$ 的最简表达式．
【难度】
【出处】
2000第15届CMO试题
【标注】
【答案】
略
【解析】
令 $b_{n}=\dfrac{a_{n}}{n !}$，则 $b_{1}=0, b_{2}=\dfrac{1}{2}, b_{3}=\dfrac{1}{3},b_{n}=\dfrac{1}{2}\left(b_{n-1}+b_{n-2}\right)+(-1)^{n} \dfrac{1-\dfrac{n}{2}}{n !}$ 令 $\begin{aligned} g_{n}=& \dfrac{1}{n !} f_{n}=\dfrac{1}{n !} \sum_{k=1}^{n}(n-k+1) \mathrm{C}_{n}^{n-k} a_{k}= \sum_{k=1}^{n} \dfrac{n-k+1}{(n-k) !} b_{k} \end{aligned}$ 所以
$\displaystyle g_{n+1}-g_{n}=\sum\limits_{k=1}^{n+1} \dfrac{n-k+2}{(n+1-k) !} b_{k}-\sum_{k=1}^{n} \dfrac{n-k+1}{(n-k) !} b_{k}=$
$\displaystyle \sum\limits_{k=2}^{n+1} \dfrac{n-k+2}{(n+1-k) !} b_{k}-\sum_{k=1}^{n+1} \dfrac{n-k+2}{(n-k+1) !} b_{k-1}=\sum_{k=2}^{n+1} \dfrac{n-k+2}{(n+1-k) !}\left(b_{k}-b_{k-1}\right)$
但是令 $d_{n}=(-2)^{n}\left(b_{n}-b_{n-1}\right)$，则 $d_{n}=d_{n-1}+\left(1-\dfrac{n}{2}\right) \dfrac{2^{n}}{n !}$
所以 $d_{2}=2,d_{n}=2+\sum_{l=3}^{n}\left(1-\dfrac{l}{2}\right) \dfrac{2^{l}}{l !}=\dfrac{2^{n}}{n !}$ 故 $b_{n}-b_{n-1}=\dfrac{d_{n}}{(-2)^{n}}=\dfrac{2^{n}}{(-2)^{n} n !}=\dfrac{(-1)^{n}}{n !}$ 所以
$\begin{aligned} g_{n+1}-g_{n}=& \sum_{k=2}^{n+1} \dfrac{n+2-k}{(n+1-k) !} \cdot \dfrac{(-1)^{k}}{k !}= \sum_{k=2}^{n} \dfrac{(-1)^{k}}{(n-k) ! k !}+\sum_{k=2}^{n+1} \dfrac{(-1)^{k}}{(n+1-k) ! k !}= \dfrac{1}{n !} \sum_{k=2}^{n}(-1)^{k}\left(\begin{array}{l}{n} \\ {k}\end{array}\right)+\dfrac{(-1)^{k}}{(n+1) !} \sum_{k=2}^{n+1}(-1)^{k} \cdot\left(\begin{array}{c}{n+1} \\ {k}\end{array}\right) \end{aligned}$
因为 $\displaystyle \sum\limits_{k=0}^{n}(-1)^{k}\left(\begin{array}{l}{n} \\ {k}\end{array}\right)=0,\sum_{k=0}^{n+1}(-1)^{k}\left(\begin{array}{c}{n+1} \\ {k}\end{array}\right)=0$
所以 $\begin{aligned} g_{n+1}-g_{n}=&-\frac{1}{n !}(1-n)-\frac{1}{(n+1) !}(1-(n+1))= \frac{1}{(n-1) !}-\frac{1}{(n+1) !} \end{aligned}$
由于 $g_{3}=2 b_{2}+b_{3}=\dfrac{4}{3}$，则
$\begin{aligned} f_{n}=& n ! g_{n}=n !\left(\sum_{k=4}^{n} \frac{1}{(k-2) !}-\sum_{k=4}^{n} \frac{1}{k !}+g_{3}\right)= n !\left(g_{3}+\frac{1}{2 !}+\frac{1}{3 !}-\frac{n+1}{n !}\right)= 2 \cdot n !-(n+1) \end{aligned}$
$\displaystyle g_{n+1}-g_{n}=\sum\limits_{k=1}^{n+1} \dfrac{n-k+2}{(n+1-k) !} b_{k}-\sum_{k=1}^{n} \dfrac{n-k+1}{(n-k) !} b_{k}=$
$\displaystyle \sum\limits_{k=2}^{n+1} \dfrac{n-k+2}{(n+1-k) !} b_{k}-\sum_{k=1}^{n+1} \dfrac{n-k+2}{(n-k+1) !} b_{k-1}=\sum_{k=2}^{n+1} \dfrac{n-k+2}{(n+1-k) !}\left(b_{k}-b_{k-1}\right)$
但是令 $d_{n}=(-2)^{n}\left(b_{n}-b_{n-1}\right)$，则 $d_{n}=d_{n-1}+\left(1-\dfrac{n}{2}\right) \dfrac{2^{n}}{n !}$
所以 $d_{2}=2,d_{n}=2+\sum_{l=3}^{n}\left(1-\dfrac{l}{2}\right) \dfrac{2^{l}}{l !}=\dfrac{2^{n}}{n !}$ 故 $b_{n}-b_{n-1}=\dfrac{d_{n}}{(-2)^{n}}=\dfrac{2^{n}}{(-2)^{n} n !}=\dfrac{(-1)^{n}}{n !}$ 所以
$\begin{aligned} g_{n+1}-g_{n}=& \sum_{k=2}^{n+1} \dfrac{n+2-k}{(n+1-k) !} \cdot \dfrac{(-1)^{k}}{k !}= \sum_{k=2}^{n} \dfrac{(-1)^{k}}{(n-k) ! k !}+\sum_{k=2}^{n+1} \dfrac{(-1)^{k}}{(n+1-k) ! k !}= \dfrac{1}{n !} \sum_{k=2}^{n}(-1)^{k}\left(\begin{array}{l}{n} \\ {k}\end{array}\right)+\dfrac{(-1)^{k}}{(n+1) !} \sum_{k=2}^{n+1}(-1)^{k} \cdot\left(\begin{array}{c}{n+1} \\ {k}\end{array}\right) \end{aligned}$
因为 $\displaystyle \sum\limits_{k=0}^{n}(-1)^{k}\left(\begin{array}{l}{n} \\ {k}\end{array}\right)=0,\sum_{k=0}^{n+1}(-1)^{k}\left(\begin{array}{c}{n+1} \\ {k}\end{array}\right)=0$
所以 $\begin{aligned} g_{n+1}-g_{n}=&-\frac{1}{n !}(1-n)-\frac{1}{(n+1) !}(1-(n+1))= \frac{1}{(n-1) !}-\frac{1}{(n+1) !} \end{aligned}$
由于 $g_{3}=2 b_{2}+b_{3}=\dfrac{4}{3}$，则
$\begin{aligned} f_{n}=& n ! g_{n}=n !\left(\sum_{k=4}^{n} \frac{1}{(k-2) !}-\sum_{k=4}^{n} \frac{1}{k !}+g_{3}\right)= n !\left(g_{3}+\frac{1}{2 !}+\frac{1}{3 !}-\frac{n+1}{n !}\right)= 2 \cdot n !-(n+1) \end{aligned}$
答案
解析
备注
