Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 5"
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Revision as of 14:29, 3 April 2012
Problem
Given that and
find
.
Solution
Multiplying both sides of the equation by , we get

and subtracting the original equation from this one we get

Using the formula for an infinite geometric series, we find

Rearranging, we get

Thus , and the answer is
.