25~26高三上·江西南昌期末·第14题

$2^n + n^2 + n$

【题目】已知数列${a_n}$满足$a_{n+1}=2a_n-n^2+n+2$，且$a_1=4$，则$a_n=$______.

【答案】$2^n + n^2 + n$

【解析】设$a_{n+1}+x(n+1)^2+y(n+1)+z=2(a_n+xn^2+yn+z)$，则$a_{n+1}=2a_n+xn^2+(y-2x)n-x-y+z$.

由$\begin{cases} x = -1 \ y – 2x = 1 \ -x – y + z = 2 \end{cases}$，解得$\begin{cases} x = -1 \ y = -1 \ z = 0 \end{cases}$.

$\therefore a_{n+1}-(n+1)^2-(n+1)=2(a_n-n^2-n)$.

又$a_1-1^2-1=2$，所以数列${a_n-n^2-n}$是以$2$为首项，$2$为公比的等比数列.

$\therefore a_n-n^2-n=2\cdot2^{n-1}=2^n$，$\therefore a_n=2^n+n^2+n$.

故答案为：$2^n + n^2 + n$.