AIME II 2005 P3

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Problem Statement

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

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Solution

$$\text{Let } a, a > 0 \text{ be the first term of the original infinite geometric series}$$ $$\text{Let } r, \vert r \vert < 1 \text{ be the common ratio of the original infinite geometric series}$$ $$\implies \dfrac{a}{1-r} = 2005, \dfrac{a^2}{1 - r^2} = 10(2005)$$ $$\dfrac{a^2}{1 - r^2} = (10)(\dfrac{a}{1-r})$$ $$\dfrac{a^2}{1 + r} = 10a$$ $$a = 10(1 + r)$$ $$\Sub a = 10(1 + r) into \dfrac{a}{1-r} = 2005$$ $$\dfrac{10(1+r)}{1-r} = 2005$$ $$10 + 10r = 2005 - 2005r \implies 2015r = 1995 \implies r = \dfrac{1995}{2015} = \dfrac{399}{403}$$ $$\text{let } \dfrac{m}{n} = \dfrac{399}{403} \implies m + n = 399 + 403 = 802$$