91学 首页 > 数学 > 高考题 > 2022 > 2022年上海春 > 正文 返回 打印

2022年高考数学上海春18

  2022-12-16 20:50:43  

(14分)已知在数列$\{a_{n}\}$中,$a_{2}=1$,其前$n$项和为$S_{n}$.
(1)若$\{a_{n}\}$是等比数列,$S_{2}=3$,求$\mathop{\lim}\limits_{n\rightarrow \infty }S_{n}$;
(2)若$\{a_{n}\}$是等差数列,$S_{2n}\geqslant n$,求其公差$d$的取值范围.
分析:(1)由已知求得等比数列的公比,再求出前$n$项和,求极限得答案;
(2)求出等差数列的前$2n$项和,代入$S_{2n}\geqslant n$,对$n$分类分析得答案.
解:(1)在等比数列$\{a_{n}\}$中,$a_{2}=1$,$S_{2}=3$,则$a_{1}=2$,
$\therefore$公比$q=\dfrac{1}{2}$,则${S}_{n}=\dfrac{{a}_{1}(1-{q}^{n})}{1-q}=4(1-\dfrac{1}{{2}^{n}})$,
$\therefore$$\mathop{\lim}\limits_{n\rightarrow \infty }S_{n}=\lim\limits_{n\rightarrow \infty }4(1-\dfrac{1}{{2}^{n}})=4$;
(2)若$\{a_{n}\}$是等差数列,
则${S}_{2n}=\dfrac{({a}_{2}+{a}_{2n-1})\cdot 2n}{2}=2d{n}^{2}+(2-3d)n\geqslant n$,
即$(3-2n)d\leqslant 1$,当$n=1$时,$d\leqslant 1$;
当$n\geqslant 2$时,$d\geqslant \dfrac{1}{3-2n}$恒成立,$\because$$\dfrac{1}{3-2n}\in [-1$,$0)$,$\therefore d\geqslant 0$.
综上所述,$d\in [0$,$1]$.
点评:本题考查等差数列与等比数列前$n$项和,考查数列极限的求法,考查数列的函数特性及应用,是中档题.

http://x.91apu.com//shuxue/gkt/2022/2022shc/2022-12-16/33615.html