Question

一个无穷等比级数的和为 $2005$ 。将原级数的每一项平方后得到一个新级数，其和是原级数和的 $10$ 倍。原级数的公比为 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数。求 $m + n$ 。

An infinite geometric series has sum $2005.$ A new series, obtained by squaring each term of the original series, has sum $10$ times the sum of the original series. The common ratio of the original series is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：802

知识点：等比数列平方差

难度评级：2070

解答：

设原级数首项为 $a$ ，公比为 $r$ ，则 $\frac{a}{1-r} = 2005$ 。平方后的级数是首项 $a^2$ 、公比 $r^2$ 的等比级数，所以 $\begin{aligned} \frac{a^2}{1-r^2} &= \frac{a}{1-r} \cdot \frac{a}{1+r} \\ &= 2005 \cdot \frac{a}{1+r} \\ &= 10 \cdot 2005, \end{aligned}$ 因此 $\frac{a}{1+r} = 10$ 。

两个方程相除，得 $\frac{1+r}{1-r} = \frac{2005}{10}$ ，所以 $2(1+r) = 401(1-r)$ ，进而 $403r = 399$ ， $r = \frac{399}{403}$ 。由于 $399 = 3 \cdot 7 \cdot 19$ 且 $403 = 13 \cdot 31$ ，该分数已最简，故 $m + n = 399 + 403 = 802$ 。

Let the original series have first term $a$ and ratio $r,$ so $\frac{a}{1-r} = 2005.$ The squared series is geometric with first term $a^2$ and ratio $r^2,$ so $\begin{aligned} \frac{a^2}{1-r^2} &= \frac{a}{1-r} \cdot \frac{a}{1+r} \\ &= 2005 \cdot \frac{a}{1+r} \\ &= 10 \cdot 2005, \end{aligned}$ which gives $\frac{a}{1+r} = 10.$

Dividing the two equations, $\frac{1+r}{1-r} = \frac{2005}{10},$ so $2(1+r) = 401(1-r),$ giving $403r = 399$ and $r = \frac{399}{403}.$ Since $399 = 3 \cdot 7 \cdot 19$ and $403 = 13 \cdot 31,$ the fraction is in lowest terms, and $m + n = 399 + 403 = 802.$

2005 AIME II 第 3 题

答案：802

难度评级：2070

解答：

其他年份的第 3 题