Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 5"
m |
|||
Line 14: | Line 14: | ||
==See also== | ==See also== | ||
− | *[[Mock AIME 2 2006-2007/Problem 4 | Previous Problem]] | + | *[[Mock AIME 2 2006-2007 Problems/Problem 4 | Previous Problem]] |
− | *[[Mock AIME 2 2006-2007/Problem 6 | Next Problem]] | + | *[[Mock AIME 2 2006-2007 Problems/Problem 6 | Next Problem]] |
*[[Mock AIME 2 2006-2007]] | *[[Mock AIME 2 2006-2007]] |
Revision as of 14:34, 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
.