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2023年高考数学新高考Ⅱ-18

  2023-07-08 11:19:52  

(12分)已知$\{a_{n}\}$为等差数列,${{b}_{n}}=\left\{ \begin{array}{*{35}{l}}
   {{a}_{n}}-6,n  \\
   2{{a}_{n}},n  \\
\end{array} \right.$,记$S_{n}$,$T_{n}$为$\{a_{n}\}$,$\{b_{n}\}$的前$n$项和,$S_{4}=32$,$T_{3}=16$.
(1)求$\{a_{n}\}$的通项公式;
(2)证明:当$n > 5$时,$T_{n} > S_{n}$.
分析:(1)根据已知条件,结合等差数列的性质,以及等差数列的前$n$项和公式,即可求解;
(2)根据已知条件,求出$T_{n}$,$S_{n}$,再结合作差法,并分类讨论,即可求证.
解:(1)设等差数列$\{a_{n}\}$的公差为$d$,
$S_{n}$,$T_{n}$为$\{a_{n}\}\{b_{n}\}$的前$n$项和,$S_{4}=32$,$T_{3}=16$,
则$\left\{\begin{array}{l}{{a}_{1}+{a}_{2}+{a}_{3}+{a}_{4}=32}\\ {{a}_{1}-6+2{a}_{2}+{a}_{3}-6=16}\end{array}\right.$,即$\left\{\begin{array}{l}{4{a}_{1}+\dfrac{4(4-1)}{2}d=32}\\ {{a}_{2}=7}\end{array}\right.$,解得$\left\{\begin{array}{l}{{a}_{1}=5}\\ {d=2}\end{array}\right.$,
故$a_{n}=5+2(n-1)=2n+3$;
(2)证明:由(1)可知,${{b}_{n}}=\left\{ \begin{array}{*{35}{l}}
   2n-3,n  \\
   4n+6,n  \\
\end{array} \right.$,
$S_n=\dfrac{(5+2n+3)n}{2}=(n+4)n$,
当$n$为偶数时,$n > 5$,
$T_{n}=-1+3+\cdot \cdot \cdot +2(n-1)-3+14+22+\cdot \cdot \cdot +4n+6$
$=\dfrac{\dfrac{n}{2}[-1+2(n-1)-3]}{2}+\dfrac{\dfrac{n}{2}(14+4n+6)}{2}=\dfrac{\dfrac{n}{2}(14+6n)}{2}=\dfrac{n(3n+7)}{2}$,
${T}_{n}-{S}_{n}=\dfrac{{n}^{2}-n}{2} > 0$,
当$n$为奇数时,$n > 5$,$T_{n}=T_{n-1}+b_{n}=\dfrac{(n-1)(3n+4)}{2}+2n-3=\dfrac{3{n}^{2}+5n-10}{2}$,
$T_{n}-S_{n}=\dfrac{{n}^{2}-3n-10}{2} > \dfrac{25-15-10}{2}=0$,
故原式得证.
点评:本题主要考查数列的求和,考查转化能力,属于中档题.

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