《2025年高考总复习首选用卷数学人教版》

16.(本小题满分15分)已知数列$\{ a_{n}\} $满足$a_{1}=2,a_{n + 1}=\frac{1}{3}a_{n}+2n+\frac{5}{3}$.
(1)求$\{ a_{n}\} $的通项公式;
(2)求数列$\{ a_{n}\} $的前$n$项和$S_{n}$.

答案： 解
(1)$a_{n + 1}=\frac{1}{3}a_{n}+2n+\frac{5}{3}$，设$a_{n + 1}+A(n + 1)+B=\frac{1}{3}(a_{n}+An + B)$，即$a_{n + 1}=\frac{1}{3}a_{n}-\frac{2}{3}An - A-\frac{2}{3}B$，即$\begin{cases}-\frac{2}{3}A = 2,\\-A-\frac{2}{3}B=\frac{5}{3},\end{cases}$解得$\begin{cases}A=-3,\\B = 2,\end{cases}$ $a_{1}-3 + 2 = 1$，故$\{ a_{n}-3n + 2\}$是首项为1，公比为$\frac{1}{3}$的等比数列. $a_{n}-3n + 2=(\frac{1}{3})^{n - 1}$，故$a_{n}=(\frac{1}{3})^{n - 1}+3n - 2$.
(2)$a_{n}=(\frac{1}{3})^{n - 1}+3n - 2$，则$S_{n}=(\frac{1}{3})^{0}+1+(\frac{1}{3})^{1}+4+\cdots+(\frac{1}{3})^{n - 1}+3n - 2=[(\frac{1}{3})^{0}+(\frac{1}{3})^{1}+\cdots+(\frac{1}{3})^{n - 1}]+(1 + 4+\cdots+3n - 2)=1\times\frac{1-(\frac{1}{3})^{n}}{1-\frac{1}{3}}+\frac{(1 + 3n - 2)\times n}{2}=-\frac{3}{2}\times(\frac{1}{3})^{n}+\frac{3}{2}n^{2}-\frac{1}{2}n+\frac{3}{2}$.