题目

已知数列 $\{a_{n}\}$ 中 $a_{1}=\dfrac{1}{2}$，且 $a_{n+1}=\dfrac{1}{2}a_{n}+\dfrac{2n+3}{2^{n+1}},n\in\mathbb N^{*}$．

【难度】

【出处】

2008年全国高中数学联赛河南省预赛

【标注】

求数列 $\{a_{n}\}$ 的通项公式；
标注
答案

$a_{n}=\dfrac{n^{2}+2n-2}{2^{n}}$

解析

因为 $a_{n+1}=\dfrac{1}{2}a_{n}+\dfrac{2n+3}{2^{n+1}}$，所以\[2^{n+1}a_{n+1}=2^{n}a_{n}+2n+3.\]令 $b_{n}=2^{n}a_{n}$，则$$b_{1}=1 , b_{n+1}=b_{n}+2n+3,$$所以\[\begin{split}b_{n}&=(b_{n}-b_{n-1})+(b_{n-1}-b_{n-2})+\cdots+(b_{2}-b_{1})+b_{1}\\&=(2n+1)+(2n-1)+\cdots+5+1\\&=\dfrac{1+(2n+1)}{2}(n+1)-3\\&=n^{2}+2n-2,\end{split}\]故 $a_{n}=\dfrac{n^{2}+2n-2}{2^{n}}$．
求数列 $\{a_{n}\}$ 的前 $n$ 项和 $S_{n}$．
标注
答案

$S_{n}=8-\dfrac{n^{2}+6n+8}{2^{n}}$

解析

由 $(1)$ 知$$\begin{split}\displaystyle S_{n}&=\sum\limits_{k=1}^{n}a_{k}\\&=\sum\limits_{k=1}^{n}\dfrac{k^{2}+2k-2}{2^{k}}\\&=\sum\limits_{k=1}^{n}\dfrac{k^{2}}{2^{k}}+2\sum\limits_{k=1}^{n}\dfrac{k}{2^{k}}-2\sum\limits_{k=1}^{n}\dfrac{1}{2^{k}}.\end{split}$$令 $\displaystyle P_{n}=\sum\limits_{k=1}^{n}\dfrac{k^{2}}{2^{k}}$，则\[\dfrac{1}{2}P_{n}=P_{n}-\dfrac{1}{2}P_{n}=\sum\limits_{k=1}^{n}\dfrac{k^{2}}{2^{k}}-\dfrac{1}{2}\sum\limits_{k=1}^{n}\dfrac{k^{2}}{2^{k}}=\sum\limits_{k=1}^{n}\dfrac{2k-1}{2^{k}}-\dfrac{n^{2}}{2^{n+1}},\]所以\[\dfrac{1}{2}P_{n}=\sum\limits_{k=1}^{n}\dfrac{2k-1}{2^{k}}-\dfrac{n^{2}}{2^{n+1}},\]因此\[\begin{split}\dfrac{1}{4}P_{n}&=\dfrac{1}{2}P_{n}-\dfrac{1}{4}P_{n}\\&=\dfrac{1}{2}+\dfrac{2}{2^{2}}+\dfrac{2}{2^{3}}+\cdots+\dfrac{2}{2^{n}}-\dfrac{2n-1}{2^{n+1}}-\dfrac{n^{2}}{2^{n+2}}\\&=\dfrac{3}{2}-\dfrac{1}{2^{n-1}}-\dfrac{2n-1}{2^{n+1}}-\dfrac{n^{2}}{2^{n+2}},\end{split}\]故\[P_{n}=6-\dfrac{4}{2^{n-1}}-\dfrac{2n-1}{2^{n-1}}-\dfrac{n^{2}}{2^{n}}.\]用错位相减法得\[\sum\limits_{k=1}^{n}\dfrac{k}{2^{k}}=2-\dfrac{1}{2^{n-1}}-\dfrac{n}{2^{n}},\]而$$\displaystyle \sum\limits_{k=1}^{n}\dfrac{1}{2^{k}}=1-\dfrac{1}{2^{n}},$$因此\[\begin{split}S_{n}&=6-\dfrac{4}{2^{n-1}}-\dfrac{2n-1}{2^{n-1}}-\dfrac{n^{2}}{2^{n}}+2\left(2-\dfrac{1}{2^{n-1}}-\dfrac{n}{2^{n}}\right)-2\left(1-\dfrac{1}{2^{n}}\right)\\&=8-\dfrac{n^{2}+6n+8}{2^{n}}.\end{split}\]

题目问题1 答案1 解析1 备注1 问题2 答案2 解析2

由 $(1)$ 知$$\begin{split}\displaystyle S_{n}&=\sum\limits_{k=1}^{n}a_{k}\\&=\sum\limits_{k=1}^{n}\dfrac{k^{2}+2k-2}{2^{k}}\\&=\sum\limits_{k=1}^{n}\dfrac{k^{2}}{2^{k}}+2\sum\limits_{k=1}^{n}\dfrac{k}{2^{k}}-2\sum\limits_{k=1}^{n}\dfrac{1}{2^{k}}.\end{split}$$令 $\displaystyle P_{n}=\sum\limits_{k=1}^{n}\dfrac{k^{2}}{2^{k}}$，则\[\dfrac{1}{2}P_{n}=P_{n}-\dfrac{1}{2}P_{n}=\sum\limits_{k=1}^{n}\dfrac{k^{2}}{2^{k}}-\dfrac{1}{2}\sum\limits_{k=1}^{n}\dfrac{k^{2}}{2^{k}}=\sum\limits_{k=1}^{n}\dfrac{2k-1}{2^{k}}-\dfrac{n^{2}}{2^{n+1}},\]所以\[\dfrac{1}{2}P_{n}=\sum\limits_{k=1}^{n}\dfrac{2k-1}{2^{k}}-\dfrac{n^{2}}{2^{n+1}},\]因此\[\begin{split}\dfrac{1}{4}P_{n}&=\dfrac{1}{2}P_{n}-\dfrac{1}{4}P_{n}\\&=\dfrac{1}{2}+\dfrac{2}{2^{2}}+\dfrac{2}{2^{3}}+\cdots+\dfrac{2}{2^{n}}-\dfrac{2n-1}{2^{n+1}}-\dfrac{n^{2}}{2^{n+2}}\\&=\dfrac{3}{2}-\dfrac{1}{2^{n-1}}-\dfrac{2n-1}{2^{n+1}}-\dfrac{n^{2}}{2^{n+2}},\end{split}\]故\[P_{n}=6-\dfrac{4}{2^{n-1}}-\dfrac{2n-1}{2^{n-1}}-\dfrac{n^{2}}{2^{n}}.\]用错位相减法得\[\sum\limits_{k=1}^{n}\dfrac{k}{2^{k}}=2-\dfrac{1}{2^{n-1}}-\dfrac{n}{2^{n}},\]而$$\displaystyle \sum\limits_{k=1}^{n}\dfrac{1}{2^{k}}=1-\dfrac{1}{2^{n}},$$因此\[\begin{split}S_{n}&=6-\dfrac{4}{2^{n-1}}-\dfrac{2n-1}{2^{n-1}}-\dfrac{n^{2}}{2^{n}}+2\left(2-\dfrac{1}{2^{n-1}}-\dfrac{n}{2^{n}}\right)-2\left(1-\dfrac{1}{2^{n}}\right)\\&=8-\dfrac{n^{2}+6n+8}{2^{n}}.\end{split}\]