Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 5"

Revision as of 15:29, 3 April 2012

Given that $iz^2=1+\frac 2z + \frac{3}{z^2}+\frac{4}{z ^3}+\frac{5}{z^4}+\cdots$ and $z=n\pm \sqrt{-i},$ find $\lfloor 100n \rfloor$ .

Multiplying both sides of the equation by $z$ , 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 $n=1$ , and the answer is $\lfloor 100n \rfloor = \boxed{100}$ .

Revision as of 02:35, 23 April 2008 (view source) I like pie (talk \| contribs) ← Older edit	Revision as of 15:29, 3 April 2012 (view source) 1=2 (talk \| contribs) m (moved Mock AIME 2 2006-2007/Problem 5 to Mock AIME 2 2006-2007 Problems/Problem 5) Newer edit →
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