题目

设 $\{a_n\}$ 是正项递增的等差数列，求证：

【难度】

【出处】

2012年全国高中数学联赛湖南省预赛

【标注】

对任意的 $k,l\in\mathbb N^*$，当 $l>k\geqslant2$ 时，$\dfrac{a_{l+1}}{a_{k+1}}<\dfrac{a_l}{a_k}<\dfrac{a_{l-1}}{a_{k-1}}$；
标注
答案

略

解析

因为 $a_l>a_k>0$，公差 $d>0$，故$$\dfrac{a_{l+1}}{a_{k+1}}=\dfrac{a_l+d}{a_k+d}<\dfrac{a_l}{a_k}<\dfrac{a_l-d}{a_k-d}=\dfrac{a_{l-1}}{a_{k-1}}.$$
对任意的 $k\in\mathbb N^*$，当 $k\geqslant2$ 时，$$\sqrt[k]{\dfrac{a_{2013k+1}}{a_{k+1}}}<\dfrac{a_{k+2}}{a_{k+1}}\cdot\dfrac{a_{2k+2}}{a_{2k+1}}\cdot\dfrac{a_{3k+2}}{a_{3k+1}}\cdots\dfrac{a_{2012k+2}}{a_{2012k+1}}<\sqrt[k]{\dfrac{a_{2012k+2}}{a_2}}.$$
标注
答案

略

解析

由 $(1)$ 知，$$\dfrac{a_{2012k+2}}{a_{2012k+1}}<\dfrac{a_{2012k+1}}{a_{2012k}}<\dfrac{a_{2012k}}{a_{2012k-1}}.$$设 $A=\dfrac{a_{k+2}}{a_{k+1}}\cdot\dfrac{a_{2k+2}}{a_{2k+1}}\cdot\dfrac{a_{3k+2}}{a_{3k+1}}\cdots\dfrac{a_{2012k+2}}{a_{2012k+1}}$，则\[\begin{split}A^k&=\left(\dfrac{a_{k+2}}{a_{k+1}}\right)^k\cdot\left(\dfrac{a_{2k+2}}{a_{2k+1}}\right)^k\cdots\left(\dfrac{a_{2012k+2}}{a_{2012k+1}}\right)^k\\&>\left(\dfrac{a_{k+2}}{a_{k+1}}\cdot\dfrac{a_{k+3}}{a_{k+2}}\cdots\dfrac{a_{2k+1}}{a_{2k}}\right)\left(\dfrac{a_{2k+2}}{a_{2k+1}}\cdot\dfrac{a_{2k+3}}{a_{2k+2}}\cdots\dfrac{a_{3k+1}}{a_{3k}}\right)\cdots\left(\dfrac{a_{2012k+2}}{a_{2012k+1}}\cdot\dfrac{a_{2012k+3}}{a_{2012k+2}}\cdots\dfrac{a_{2013k+1}}{a_{2013k}}\right)\\&=\dfrac{a_{2013k+1}}{a_{k+1}}.\end{split}\]又因为\[\begin{split}A^k&=\left(\dfrac{a_{k+2}}{a_{k+1}}\right)^k\cdot\left(\dfrac{a_{2k+2}}{a_{2k+1}}\right)^k\cdots\left(\dfrac{a_{2012k+2}}{a_{2012k+1}}\right)^k\\&<\left(\dfrac{a_3}{a_2}\cdot\dfrac{a_4}{a_3}\cdots\dfrac{a_{k+2}}{a_{k+1}}\right)\left(\dfrac{a_{k+3}}{a_{k+2}}\cdot\dfrac{a_{k+4}}{a_{k+3}}\cdots\dfrac{a_{2k+2}}{a_{2k+1}}\right)\cdots\left(\dfrac{a_{2011k+3}}{a_{2011k+2}}\cdot\dfrac{a_{2011k+4}}{a_{2011k+3}}\cdots\dfrac{a_{2012k+2}}{a_{2012k+1}}\right)\\&=\dfrac{a_{2012k+2}}{a_2},\end{split}\]因此$$\sqrt[k]{\dfrac{a_{2013k+1}}{a_{k+1}}}<\dfrac{a_{k+2}}{a_{k+1}}\cdot\dfrac{a_{2k+2}}{a_{2k+1}}\cdot\dfrac{a_{3k+2}}{a_{3k+1}}\cdots\dfrac{a_{2012k+2}}{a_{2012k+1}}<\sqrt[k]{\dfrac{a_{2012k+2}}{a_2}}.$$

题目问题1 答案1 解析1 备注1 问题2 答案2 解析2

由 $(1)$ 知，$$\dfrac{a_{2012k+2}}{a_{2012k+1}}<\dfrac{a_{2012k+1}}{a_{2012k}}<\dfrac{a_{2012k}}{a_{2012k-1}}.$$设 $A=\dfrac{a_{k+2}}{a_{k+1}}\cdot\dfrac{a_{2k+2}}{a_{2k+1}}\cdot\dfrac{a_{3k+2}}{a_{3k+1}}\cdots\dfrac{a_{2012k+2}}{a_{2012k+1}}$，则\[\begin{split}A^k&=\left(\dfrac{a_{k+2}}{a_{k+1}}\right)^k\cdot\left(\dfrac{a_{2k+2}}{a_{2k+1}}\right)^k\cdots\left(\dfrac{a_{2012k+2}}{a_{2012k+1}}\right)^k\\&>\left(\dfrac{a_{k+2}}{a_{k+1}}\cdot\dfrac{a_{k+3}}{a_{k+2}}\cdots\dfrac{a_{2k+1}}{a_{2k}}\right)\left(\dfrac{a_{2k+2}}{a_{2k+1}}\cdot\dfrac{a_{2k+3}}{a_{2k+2}}\cdots\dfrac{a_{3k+1}}{a_{3k}}\right)\cdots\left(\dfrac{a_{2012k+2}}{a_{2012k+1}}\cdot\dfrac{a_{2012k+3}}{a_{2012k+2}}\cdots\dfrac{a_{2013k+1}}{a_{2013k}}\right)\\&=\dfrac{a_{2013k+1}}{a_{k+1}}.\end{split}\]又因为\[\begin{split}A^k&=\left(\dfrac{a_{k+2}}{a_{k+1}}\right)^k\cdot\left(\dfrac{a_{2k+2}}{a_{2k+1}}\right)^k\cdots\left(\dfrac{a_{2012k+2}}{a_{2012k+1}}\right)^k\\&<\left(\dfrac{a_3}{a_2}\cdot\dfrac{a_4}{a_3}\cdots\dfrac{a_{k+2}}{a_{k+1}}\right)\left(\dfrac{a_{k+3}}{a_{k+2}}\cdot\dfrac{a_{k+4}}{a_{k+3}}\cdots\dfrac{a_{2k+2}}{a_{2k+1}}\right)\cdots\left(\dfrac{a_{2011k+3}}{a_{2011k+2}}\cdot\dfrac{a_{2011k+4}}{a_{2011k+3}}\cdots\dfrac{a_{2012k+2}}{a_{2012k+1}}\right)\\&=\dfrac{a_{2012k+2}}{a_2},\end{split}\]因此$$\sqrt[k]{\dfrac{a_{2013k+1}}{a_{k+1}}}<\dfrac{a_{k+2}}{a_{k+1}}\cdot\dfrac{a_{2k+2}}{a_{2k+1}}\cdot\dfrac{a_{3k+2}}{a_{3k+1}}\cdots\dfrac{a_{2012k+2}}{a_{2012k+1}}<\sqrt[k]{\dfrac{a_{2012k+2}}{a_2}}.$$