每日一题[1006]通项与和

已知数列$\{a_n\}$的前$n$项和为$S_n$,对任意的$n\in\mathbb N^*$,$S_n=(-1)^na_n+\dfrac{1}{2^n}+n-3$且$(a_{n+1}-p)(a_n-p)<0$恒成立,则实数$p$的取值范围是_______.


cover

正确答案是$\left(-\dfrac 34,\dfrac {11}4\right)$.

分析与解 在题中等式中分别令$n=2k-1,2k,2k+1$,$k\in\mathbb N^*$,有\[\begin{aligned}S_{2k-1}&=-a_{2k-1}+\dfrac{1}{2^{2k-1}}+2k-4,\\S_{2k}&=a_{2k}+\dfrac{1}{2^{2k}}+2k-3,\\S_{2k+1}&=-a_{2k+1}+\dfrac{1}{2^{2k+1}}+2k-2,\end{aligned}\]于是\[\begin{aligned}a_{2k}&=a_{2k}+a_{2k-1}-\dfrac{1}{2^{2k}}+1,\\a_{2k+1}&=-a_{2k+1}-a_{2k}-\dfrac{1}{2^{2k+1}}+1,\end{aligned}\]进而可得\[\begin{aligned}a_{2k-1}&=\dfrac{1}{4^k}-1,\\
a_{2k}&=3-\dfrac{1}{4^k}.\end{aligned}\]

接下来考虑$p$的取值范围.根据题意,$p$在数列$\{a_n\}$的任意相邻两项之间.

一方面,有$a_1<p<a_2$,即$-\dfrac 34<p<\dfrac{11}4$.

另一方面,当$-\dfrac 34<p<\dfrac{11}4$时,有\[a_{2k-1}-p<a_1-p<0,\]且\[a_{2k}-p>a_2-p>0,\]于是有\[\forall n\in\mathbb N^*,(a_{n+1}-p)(a_n-p)<0.\]

综上所述,实数$p$的取值范围是$\left(-\dfrac 34,\dfrac {11}4\right)$.

此条目发表在每日一题分类目录,贴了标签。将固定链接加入收藏夹。

发表评论