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_______．

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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$\left(-\dfrac 34,\dfrac {11}4\right)$．