Question

一个正整数数列满足 $a_1 = 1$ 且 $a_9 + a_{10} = 646$ ，其构造方式为：前三项成等比数列，第二、三、四项成等差数列；一般地，对所有 $n \ge 1$ ， $a_{2n-1}$ 、 $a_{2n}$ 、 $a_{2n+1}$ 成等比数列，而 $a_{2n}$ 、 $a_{2n+1}$ 、 $a_{2n+2}$ 成等差数列。令 $a_n$ 为该数列中小于 $1000$ 的最大项。求 $n + a_n$ 。

A sequence of positive integers with $a_1 = 1$ and $a_9 + a_{10} = 646$ is formed so that the first three terms are in geometric progression, the second, third, and fourth terms are in arithmetic progression, and, in general, for all $n \ge 1,$ the terms $a_{2n-1},$ $a_{2n},$ and $a_{2n+1}$ are in geometric progression, and the terms $a_{2n},$ $a_{2n+1},$ and $a_{2n+2}$ are in arithmetic progression. Let $a_n$ be the greatest term in this sequence that is less than $1000.$ Find $n + a_n.$

答案

答案：973

知识点：等比数列等差数列找规律

难度评级：2840

小提示：

令 $a_2 = r$ 。每个等比或等差条件都会确定下一项，从而得到 $a_9 = (4r-3)^2$ 和 $a_{10} = (4r-3)(5r-4)$ 。

Let $a_2 = r;$ each progression condition forces the next term, giving $a_9 = (4r-3)^2$ and $a_{10} = (4r-3)(5r-4)$

大提示：

因而 $(4r-3)(9r-7) = 646$ ，推出 $r = 5$ ；奇数下标项为 $(2n-1)^2$ ，偶数下标项为 $(2n-3)(2n+1)$ 。

Then $(4r-3)(9r-7) = 646$ forces $r = 5,$ so odd-indexed terms are $(2n-1)^2$ and even-indexed terms are $(2n-3)(2n+1)$

解答：

令 $a_2 = r$ 。等比条件给出 $a_3 = r^2$ ，等差条件给出 $a_4 = 2r^2 - r = r(2r-1)$ ，接着 $a_5 = (2r-1)^2$ ，依此类推可归纳得到 $\begin{aligned} a_{2k+1} &= \bigl(kr - (k-1)\bigr)^2, \\ a_{2k+2} &= \bigl(kr - (k-1)\bigr) \\ &\quad {}\cdot \bigl((k+1)r - k\bigr) \end{aligned}\text{。}$ 特别地， $a_9 = (4r-3)^2$ ， $a_{10} = (4r-3)(5r-4)$ ，所以 $a_9 + a_{10} = (4r-3)(9r-7)$ $= 646$ 。展开得 $36r^2 - 55r - 625 = 0$ ，分解为 $(r - 5)(36r + 125) = 0$ ，所以 $r = 5$ 。

当 $r = 5$ 时， $kr - (k-1) = 4k + 1$ ，所以 $a_{2k+1} = (4k+1)^2$ ， $a_{2k+2} = (4k+1)(4k+5)$ ，数列递增。因为 $a_{17} = 33^2 = 1089 \gt 1000$ ，而 $a_{16} = 29 \cdot 33 = 957$ ，所以小于 $1000$ 的最大项是 $a_{16} = 957$ 。

因此 $n + a_n = 16 + 957 = 973$ 。

Let $a_2 = r.$ The geometric condition gives $a_3 = r^2,$ the arithmetic condition gives $a_4 = 2r^2 - r = r(2r-1),$ then $a_5 = (2r-1)^2,$ and so on: inductively $\begin{aligned} a_{2k+1} &= \bigl(kr - (k-1)\bigr)^2, \\ a_{2k+2} &= \bigl(kr - (k-1)\bigr) \\ &\quad {}\cdot \bigl((k+1)r - k\bigr). \end{aligned}$ In particular $a_9 = (4r-3)^2$ and $a_{10} = (4r-3)(5r-4),$ so $a_9 + a_{10} = (4r-3)(9r-7)$ $= 646.$ Expanding gives $36r^2 - 55r - 625 = 0,$ which factors as $(r - 5)(36r + 125) = 0,$ so $r = 5.$

With $r = 5$ we get $kr - (k-1) = 4k + 1,$ so $a_{2k+1} = (4k+1)^2$ and $a_{2k+2} = (4k+1)(4k+5);$ the sequence is increasing. Since $a_{17} = 33^2 = 1089 \gt 1000$ while $a_{16} = 29 \cdot 33 = 957,$ the greatest term below $1000$ is $a_{16} = 957.$

Therefore $n + a_n = 16 + 957 = 973.$

2004 AIME II 第 9 题

答案：973

难度评级：2840

小提示：

大提示：

解答：

其他年份的第 9 题