Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 5"
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Latest revision as of 09:51, 4 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
.
See also
Mock AIME 2 2006-2007 (Problems, Source) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |