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2021年高考数学乙卷-文19

  2022-05-03 08:12:49  

19.(12分)设$\{a_{n}\}$是首项为1的等比数列,数列$\{b_{n}\}$满足$b_{n}=\dfrac{n{a}_{n}}{3}$,已知$a_{1}$,$3a_{2}$,$9a_{3}$成等差数列.
(1)求$\{a_{n}\}$和$\{b_{n}\}$的通项公式;
(2)记$S_{n}$和$T_{n}$分别为$\{a_{n}\}$和$\{b_{n}\}$的前$n$项和.证明:$T_{n}<\dfrac{{S}_{n}}{2}$.
分析:(1)根据$a_{1}$,$3a_{2}$,$9a_{3}$成等差数列,$\{a_{n}\}$是首项为1的等比数列,求出公比$q$,进一步求出$\{a_{n}\}$和$\{b_{n}\}$的通项公式;
(2)分别利用等比数列的前$n$项和公式和错位相减法,求出$S_{n}$和$T_{n}$,再利用作差法证明$T_{n}<\dfrac{{S}_{n}}{2}$.
解:(1)$\because a_{1}$,$3a_{2}$,$9a_{3}$成等差数列,$\therefore 6a_{2}=a_{1}+9a_{3}$,
$\because \{a_{n}\}$是首项为1的等比数列,设其公比为$q$,
则$6q=1+9q^{2}$,$\therefore q=\dfrac{1}{3}$,
$\therefore a_{n}=a_{1}q^{n-1}=(\dfrac{1}{3})^{n-1}$,
$\therefore b_{n}=\dfrac{n{a}_{n}}{3}=n\cdot (\dfrac{1}{3})^{n}$.
(2)证明:由(1)知$a_{n}=(\dfrac{1}{3})^{n-1}$,$b_{n}=n\cdot (\dfrac{1}{3})^{n}$,
$\therefore$${S}_{n}=\dfrac{1\times [1-(\dfrac{1}{3})^{n}]}{1-\dfrac{1}{3}}=\dfrac{3}{2}-\dfrac{1}{2}\times (\dfrac{1}{3})^{n-1}$,
${T}_{n}=1\times (\dfrac{1}{3})^{1}+2\times (\dfrac{1}{3})^{2}+\ldots +n\cdot (\dfrac{1}{3})^{n}$,①
$\therefore$$\dfrac{1}{3}{T}_{n}=1\times (\dfrac{1}{3})^{2}+2\times (\dfrac{1}{3})^{3}+\ldots +n\cdot (\dfrac{1}{3})^{n+1}$,②
①$-$②得,$\dfrac{2}{3}{T}_{n}=\dfrac{1}{2}[1-(\dfrac{1}{3})^{n}]-n(\dfrac{1}{3})^{n+1}$,
$\therefore$${T}_{n}=\dfrac{3}{4}-\dfrac{1}{4}\times (\dfrac{1}{3})^{n-1}-\dfrac{n}{2}(\dfrac{1}{3})^{n}$,
$\therefore$${T}_{n}-\dfrac{{S}_{n}}{2}=\dfrac{3}{4}-\dfrac{1}{4}\times (\dfrac{1}{3})^{n-1}-\dfrac{n}{2}\cdot (\dfrac{1}{3})^{n}-[\dfrac{3}{4}-\dfrac{1}{4}\times (\dfrac{1}{3})^{n-1}]<0$,
$\therefore T_{n}<\dfrac{{S}_{n}}{2}$.
点评:本题考查了等差数列与等比数列的性质,等比数列的前$n$项和公式和利用错位相减法求数列的前$n$项和,考查了方程思想和转化思想,属中档题.

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