设数列 $\{a_n\}$ 满足：$a_1=1,a_{n+1}=2a_n+n\cdot(1+2^n),n=1,2,3,\cdots$．试求通项 $a_n$ 的表达式．
【难度】
【出处】
2008中国东南数学奥林匹克试题
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【答案】
略
【解析】
将所给递推关系的两边同时除以 $2^{n+1}$，得 $\dfrac{a_{n+1}}{2^{n+1}}=\dfrac{a_n}{2^n}+\dfrac{n}{2^{n+1}}+\dfrac{n}{2}$
即 $\dfrac{a_{n+1}}{2^{n+1}}-\dfrac{a_n}{2^n}=\dfrac{n}{2^{n+1}}+\dfrac{n}{2}$
所以 $\displaystyle \sum\limits_{i=1}^n\left(\dfrac{a_{i+1}}{2^{i+1}}-\dfrac{a_{i}}{2^{i}}\right)=\sum_{i=1}^n\dfrac{i}{2^{i+1}}+\sum_{i=1}^n\dfrac{i}{2}$
$\displaystyle \dfrac{a_{n+1}}{2^{n+1}}-\dfrac{a_1}{2^1}=\dfrac{n(n+1)}{4}+\sum\limits_{i=1}^n\dfrac{i}{2^{i+1}}$
即 $\displaystyle a_{n+1}=2^{n+1}\left[\dfrac{n(n+1)}{4}+\dfrac{1}{2^n}+\dfrac{1}{2}\sum\limits_{i=1}^n\dfrac{i}{2^i}\right]$．
令 $\displaystyle S_n=\sum\limits_{i=1}^n\dfrac{i}{2^i}$，则 $\displaystyle 2S_n=\sum\limits_{i=1}^m\dfrac{i}{2^{i-1}}$
$\begin{aligned} S_n&=2S_n-S_n=\sum_{i=1}^n\frac{i}{2^{i-1}}-\sum_{i=1}^n\frac{i}{2^i}=\sum_{i=1}^n\frac{i}{2^{i-1}}-\sum_{i=2}^{n+1}\frac{i-1}{2^{i-1}}\\
&=\frac{1}{2^{1-1}}-\frac{n+1-1}{2^{n+1-1}}+\sum_{i=2}^n\left(\dfrac{i}{2^{i-1}}-\frac{i-1}{2^{i-1}}\right)\\
&=1-\frac{n}{2^n}+\sum_{i=2}^n\frac{1}{2^{i-1}}=1-\frac{n}{2^n}+\frac{\frac{1}{2}}{1-\frac{1}{2}}\left[1-\left(\frac{1}{2}\right)^{n-1}\right]\\
&=1-\frac{n}{2^n}+1-\frac{1}{2^{n-1}}=2-\frac{n+2}{2^n}\end{aligned}$
故 $a_{n+1}=\left[\dfrac{n(n+1)}{4}+\dfrac{1}{2^n}+\dfrac{1}{2}\left(2-\dfrac{n+2}{2^n}\right)\right]=2^{n+1}\left[\dfrac{3}{2}+\dfrac{n(n+1)}{4}-\dfrac{n+2}{2^{n+1}}\right](n\geqslant 1)$
从而 $a_n=2^{n-2}(n^2-n+6)-n-1(n\geqslant 2)$．
即 $\dfrac{a_{n+1}}{2^{n+1}}-\dfrac{a_n}{2^n}=\dfrac{n}{2^{n+1}}+\dfrac{n}{2}$
所以 $\displaystyle \sum\limits_{i=1}^n\left(\dfrac{a_{i+1}}{2^{i+1}}-\dfrac{a_{i}}{2^{i}}\right)=\sum_{i=1}^n\dfrac{i}{2^{i+1}}+\sum_{i=1}^n\dfrac{i}{2}$
$\displaystyle \dfrac{a_{n+1}}{2^{n+1}}-\dfrac{a_1}{2^1}=\dfrac{n(n+1)}{4}+\sum\limits_{i=1}^n\dfrac{i}{2^{i+1}}$
即 $\displaystyle a_{n+1}=2^{n+1}\left[\dfrac{n(n+1)}{4}+\dfrac{1}{2^n}+\dfrac{1}{2}\sum\limits_{i=1}^n\dfrac{i}{2^i}\right]$．
令 $\displaystyle S_n=\sum\limits_{i=1}^n\dfrac{i}{2^i}$，则 $\displaystyle 2S_n=\sum\limits_{i=1}^m\dfrac{i}{2^{i-1}}$
$\begin{aligned} S_n&=2S_n-S_n=\sum_{i=1}^n\frac{i}{2^{i-1}}-\sum_{i=1}^n\frac{i}{2^i}=\sum_{i=1}^n\frac{i}{2^{i-1}}-\sum_{i=2}^{n+1}\frac{i-1}{2^{i-1}}\\
&=\frac{1}{2^{1-1}}-\frac{n+1-1}{2^{n+1-1}}+\sum_{i=2}^n\left(\dfrac{i}{2^{i-1}}-\frac{i-1}{2^{i-1}}\right)\\
&=1-\frac{n}{2^n}+\sum_{i=2}^n\frac{1}{2^{i-1}}=1-\frac{n}{2^n}+\frac{\frac{1}{2}}{1-\frac{1}{2}}\left[1-\left(\frac{1}{2}\right)^{n-1}\right]\\
&=1-\frac{n}{2^n}+1-\frac{1}{2^{n-1}}=2-\frac{n+2}{2^n}\end{aligned}$
故 $a_{n+1}=\left[\dfrac{n(n+1)}{4}+\dfrac{1}{2^n}+\dfrac{1}{2}\left(2-\dfrac{n+2}{2^n}\right)\right]=2^{n+1}\left[\dfrac{3}{2}+\dfrac{n(n+1)}{4}-\dfrac{n+2}{2^{n+1}}\right](n\geqslant 1)$
从而 $a_n=2^{n-2}(n^2-n+6)-n-1(n\geqslant 2)$．
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