91学 首页 > 数学 > 高考题 > 2021 > 2021年浙江 > 正文 返回 打印

2021年高考数学浙江20

  2022-05-03 08:26:47  

20.(15分)已知数列$\{a_{n}\}$的前$n$项和为$S_{n}$,$a_{1}=-\dfrac{9}{4}$,且$4S_{n+1}=3S_{n}-9(n\in N*)$.
(Ⅰ)求数列$\{a_{n}\}$的通项公式;
(Ⅱ)设数列$\{b_{n}\}$满足$3b_{n}+(n-4)a_{n}=0(n\in N*)$,记$\{b_{n}\}$的前$n$项和为$T_{n}$,若$T_{n}\leqslant \lambda b_{n}$对任意$n\in N*$恒成立,
求实数$\lambda$的取值范围.
分析:(Ⅰ)首先利用递推关系式确定数列为等比数列,然后结合等比数列的通项公式可得数列的通项公式;
(Ⅱ)首先错位相减求得$T_{n}$的值,然后分离参数利用恒成立的结论分类讨论即可求得实数$\lambda$的取值范围.
解:(Ⅰ)由$4S_{n+1}=3S_{n}?9$ 可得$4S_{n}=3S_{n?1}?9(n\geqslant 2)$,
两式作差,可得:$4a_{n+1}=3a_{n}$,
$\therefore$$\dfrac{{a}_{n+1}}{{a}_{n}}=\dfrac{3}{4}$,
很明显,$\dfrac{{a}_{2}}{{a}_{1}}=\dfrac{3}{4}$,
所以数列$\{a_{n}\}$ 是以$?\dfrac{9}{4}$为首项,$\dfrac{3}{4}$为公比的等比数列,
其通项公式为:${a}_{n}=(?\dfrac{9}{4})\times {(\dfrac{3}{4})}^{n?1}=?3\times {(\dfrac{3}{4})}^{n}$.
(Ⅱ)由$3b_{n}+(n?4)a_{n}=0$,得${b}_{n}=?\dfrac{n?4}{3}{a}_{n}=(n?4){(\dfrac{3}{4})}^{n}$,
${T}_{n}=?3\times \dfrac{3}{4}?2\times {(\dfrac{3}{4})}^{2}?1\times {(\dfrac{3}{4})}^{3}+\cdots +(n?5){(\dfrac{3}{4})}^{n?1}+(n?4)\cdot {(\dfrac{3}{4})}^{n}$,
$\dfrac{3}{4}{T}_{n}=?3\times {(\dfrac{3}{4})}^{2}?2\times {(\dfrac{3}{4})}^{3}?1\times {(\dfrac{3}{4})}^{4}+\cdots +(n?5)\cdot {(\dfrac{3}{4})}^{n}+(n?4)\cdot {(\dfrac{3}{4})}^{n+1}$,
两式作差可得:
$\dfrac{1}{4}{T}_{n}=?3\times \dfrac{3}{4}+{(\dfrac{3}{4})}^{2}+{(\dfrac{3}{4})}^{3}+{(\dfrac{3}{4})}^{4}+\cdots {(\dfrac{3}{4})}^{n}?(n?4)\cdot {(\dfrac{3}{4})}^{n+1}$
$=?\dfrac{9}{4}+\dfrac{\dfrac{9}{16}[1?{(\dfrac{3}{4})}^{n?1}]}{1?\dfrac{3}{4}}?(n?4){(\dfrac{3}{4})}^{n+1}$
$=?\dfrac{9}{4}+\dfrac{9}{4}?4{(\dfrac{3}{4})}^{n+1}?(n?4)\cdot {(\dfrac{3}{4})}^{n+1}=?n\cdot {(\dfrac{3}{4})}^{n+1}$,
则${T}_{n}=?4n\cdot {(\dfrac{3}{4})}^{n+1}$.
据此可得$?4n\cdot {(\dfrac{3}{4})}^{n+1}\leqslant \lambda (n?4){(\dfrac{3}{4})}^{n}$ 恒成立,即$\lambda (n?4)+3n\geqslant 0$ 恒成立.
$n=4$时不等式成立;
$n<4$时,$\lambda \leqslant ?\dfrac{3n}{n?4}=?3?\dfrac{12}{n?4}$,由于$n=1$时${(?3?\dfrac{12}{n?4})}_{min}=1$,故$\lambda \leqslant 1$;
$n>4$时,$\lambda \geqslant ?\dfrac{3n}{n-4}=?3?\dfrac{12}{n?4}$,而$?3?\dfrac{12}{n?4}<?3$,故:$\lambda \geqslant ?3$;
综上可得,$\{\lambda \vert ?3\leqslant \lambda \leqslant 1\}$.
点评:本题主要考查由递推关系式求数列的通项公式的方法,错位相减求和的方法,数列中的恒成立问题,分类讨论的数学思想等知识,属于中等题.

http://x.91apu.com//shuxue/gkt/2021/2021zj/2022-05-03/33352.html