已知数列 $\{a_{n}\}$ 的前 $n$ 和为 $S_{n}\}$，且 $\dfrac{S_{n}}{a_{n}}=\dfrac{1}{2}a_{n+1}(n\in\mathbb N^{*})$，其中 $a_{1}=1$，$a_{n}\ne 0$．
【难度】
【出处】
无
【标注】
-
求数列 $\{a_{n}\}$ 的通项公式；标注答案$a_{n}=n(n\in\mathbb N^{*})$解析已知$$S_{n}=\dfrac{1}{2}a_{n}a_{n+1},$$故$$\begin{split}a_{n+1}&=S_{n+1}-S_{n}\\ &=\dfrac{1}{2}a_{n+1}a_{n+2}-\dfrac{1}{2}a_{n}a_{n+1}.\end{split}$$因为 $a_{n}\ne 0$，当然 $a_{n+1}\ne 0$，所以$$a_{n+2}-a_{n}=2(n\in \mathbb N^{*}).$$由于$$a_{1}=S_{1}=\dfrac{1}{2}a_{1}a_{2},a_{1}=1,$$故 $a_{2}=2$，于是$$\begin{split}a_{2m-1}&=1+2(m-1)=2m-1,\\a_{2m}&=2+2(m-1)=2m,\end{split}$$所以 $a_{n}=n(n\in\mathbb N^{*})$．
-
设数列 $\{b_{n}\}$ 满足 $\left(2a_{n}-1\right)\left(2^{b_{n}}-1\right)=1$，$T_{n}$ 为 $\{b_{n}\}$ 的前 $n$ 项和，求证：$2T_{n}>{\log_{2}}\left(2a_{n}+1\right),n\in\mathbb N^{*}$；标注答案略解析由 $(2a_{n}-1)(2^{b_{n}}-1)=1$，得\[(2n-1)(2^{b_{n}}-1)=1,\]因为 $2^{b_{n}}=\dfrac{2n}{2n-1}$，故$$b_{n}={\log_{2}}\dfrac{2n}{2n-1},$$从而\[\begin{split}T_{n}&=b_{1}+b_{2}+\cdots+b_{n}=\log_{2}\left(\dfrac{2}{1}\cdot \dfrac{4}{3}\cdot \dfrac{6}{5}\cdots\cdots\dfrac{2n}{2n-1}\right),\\
2T_{n}&=2\log_{2}\left(\dfrac{2}{1}\cdot \dfrac{4}{3}\cdot \dfrac{6}{5}\cdots\dfrac{2n}{2n-1}\right)=\log_{2}\left(\dfrac{2}{1}\cdot \dfrac{4}{3}\cdot\dfrac{6}{5}\cdots\dfrac{2n}{2n-1}\right)^{2},\end{split}\]因此\[\begin{split}2T_{n}-\log_{2}(2a_{n}+1)&=\log_{2}\left(\dfrac{2}{1}\cdot \dfrac{4}{3}\cdot \dfrac{6}{5}\cdots\dfrac{2n}{2n-1}\right)^{2}-\log_{2}(2n+1)\\&=\log_{2}\left(\dfrac{2}{1}\cdot \dfrac{4}{3}\cdot \dfrac{6}{5}\cdots\dfrac{2n}{2n-1}\right)^{2}+\log_{2}\dfrac{1}{2n+1}\\&=\log_{2}\left[\left(\dfrac{2}{1}\cdot \dfrac{4}{3}\cdot\dfrac{6}{5}\cdots\dfrac{2n}{2n-1}\right)^{2}\cdot \dfrac{1}{2n+1}\right].\end{split}\]设$$f(n)=\left(\dfrac{2}{1}\cdot \dfrac{4}{3}\cdot \dfrac{6}{3}\cdots\dfrac{2n}{2n-1}\right)^{2}\cdot\dfrac{1}{2n+1},$$则$$f(n+1)=\left(\dfrac{2}{1}\cdot \dfrac{4}{3}\cdot \dfrac{6}{3}\cdots\dfrac{2n}{2n-1}\cdot \dfrac{2n+2}{2n+1}\right)^{2}\cdot\dfrac{1}{2n+3},$$故\[\begin{split}\dfrac{f(n+1)}{f(n)}&=\dfrac{2n+1}{2n+3}\cdot \left(\dfrac{2n+2}{2n+1}\right)^{2}\\ &=\dfrac{(2n+2)^{2}}{(2n+3)(2n+1)}\\ &=\dfrac{4n^{2}+8n+4}{4n^{2}+8n+3}>1,\end{split}\]注意到 $f(n)>0$，所以 $f(n+1)>f(n)$．
特别地$$f(n)\geqslant f(1)=\dfrac{4}{3}>1,$$从而$$2T_{n}-\log_{2}(2a_{n}+1)=\log_{2}f(n)>0.$$所以 $2T_{n}>\log_{2}(2a_{n}+1)$，$n\in\mathbb N^{*}$． -
是否存在正整数 $m,d$，使得 $\lim\limits_{n\to \infty}\left[\left(\dfrac{1}{3}\right)^{m}+\left(\dfrac{1}{3}\right)^{m+d}+\left(\dfrac{1}{3}\right)^{m+2d}+\cdots+\left(\dfrac{1}{3}\right)^{m+(n-1)d}\right]=\dfrac{1}{a_{8}}$ 成立？若存在，请求出 $m$ 和 $d$ 的值；若不存在，请说明理由．标注答案略解析易得$$\left(\dfrac{1}{3}\right)^{a_{n}}=\left(\dfrac{1}{3}\right)^{n}.$$注意到 $a_{3}=8$，则有\[\lim\limits_{n\to \infty}\left[\left(\dfrac{1}{3}\right)^{m}+\left(\dfrac{1}{3}\right)^{m+d}+\left(\dfrac{1}{3}\right)^{m+2d}+\cdots+\left(\dfrac{1}{3}\right)^{m+(n-1)d}\right]=\dfrac{\left(\dfrac{1}{3}\right)^{m}}{1-\left(\dfrac{1}{3}\right)^{d}}=\dfrac{1}{8}.\]即$$\left(\dfrac{1}{3}\right)^{m}=\dfrac{1}{8}\left[1-\left(\dfrac{1}{8}\right)^{d}\right],$$整理得\[3^{m}-3^{m-d}=8\cdots\text{ ① }\]
情形一 当 $m\geqslant d$ 时，由 ① 得$$3^{m-d}(3^{d}-1)=8.$$因为 $m,d\in\mathbb N^{*}$，所以$$m=d=2.$$情形二 当 $m<d$ 时，由 ① 得\[3^{d}-1=8\cdot 3^{d-m}\cdots\text{ ② }\]因为 $m<d$，故 ② 式右边必是 $3$ 的倍数，而左边不是 $3$ 的倍数，所以 ② 式不成立．
因此当 $m<d$ 时，不存在 $m,d\in\mathbb N^{*}$，使得 ① 式成立．
综上所述，存在正整数$$m=d=2,$$使得 $\lim\limits_{n\to \infty}\left[\left(\dfrac{1}{3}\right)^{m}+\left(\dfrac{1}{3}\right)^{m+d}+\left(\dfrac{1}{3}\right)^{m+2d}+\cdots+\left(\dfrac{1}{3}\right)^{m+(n-1)d}\right]=\dfrac{1}{a_{8}}$ 成立．
题目
问题1
答案1
解析1
备注1
问题2
答案2
解析2
备注2
问题3
答案3
解析3
备注3
