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Mock AIME 2 2006-2007/Problem 5

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Problem

Given that iz^2=1+\frac 2z + \frac{3}{z^2}+\frac{4}{z ^3}+\frac{5}{z^4}+\cdots and find .

Solution

Multiplying both sides of the equation by , we get

iz^3 = z + 2 + \frac{3}{z} + \frac{4}{z^2} + \cdots,

and subtracting the original equation from this one we get

iz^2(z-1)=z+1+\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}+\cdots.

Using the formula for an infinite geometric series, we find

iz^2(z-1)=\frac{z}{1-\frac{1}{z}}=\frac{z^2}{z-1}.

Rearranging, we get

iz^2(z-1)^2=z^2\iff (z-1)^2=\frac{1}{i}=-i\Rightarrow z=1\pm\sqrt{-i}.

Thus , and the answer is \lfloor 100n \rfloor = \boxed{100}.

See also

Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.
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