Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 5"
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==Solution== | ==Solution== | ||
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+ | Multiplying both sides of the equation by <math>z</math>, we get <P><center><math>iz^3 = z + 2 + \frac{3}{z} + \frac{4}{z^2} + \cdots,</math></center></P><div align=left>and subtracting the original equation from this one we get</div><P><center><math>iz^2(z-1)=z+1+\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}+\cdots.</math></center></P><div align=left>Using the formula for an infinite geometric series, we find</div><P><center><math>iz^2(z-1)=\frac{z}{1-\frac{1}{z}}=\frac{z^2}{z-1}.</math></center></P><div align=left>Rearranging, we get</div><P><center><math>iz^2(z-1)^2=z^2\iff (z-1)^2=\frac{1}{i}=-i\Rightarrow z=1\pm\sqrt{-i}.</math></center></P><div align=left>Thus the answer is <math>n=1, \lfloor 100n \rfloor = 100</math>.</div> | ||
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Revision as of 12:43, 16 September 2006
Problem
Given that and
find
Solution
Multiplying both sides of the equation by , we get
![$iz^3 = z + 2 + \frac{3}{z} + \frac{4}{z^2} + \cdots,$](http://latex.artofproblemsolving.com/2/1/5/2155b6e2cffdba244ad78e591c3fff17541daebf.png)
and subtracting the original equation from this one we get
![$iz^2(z-1)=z+1+\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}+\cdots.$](http://latex.artofproblemsolving.com/4/4/7/447fd4976528a61f5de07c7536f8811871ed2908.png)
Using the formula for an infinite geometric series, we find
![$iz^2(z-1)=\frac{z}{1-\frac{1}{z}}=\frac{z^2}{z-1}.$](http://latex.artofproblemsolving.com/e/8/b/e8b46f4d74f174fa6150ac93cba1d6eb29244642.png)
Rearranging, we get
![$iz^2(z-1)^2=z^2\iff (z-1)^2=\frac{1}{i}=-i\Rightarrow z=1\pm\sqrt{-i}.$](http://latex.artofproblemsolving.com/b/a/3/ba3a02366dc8eb145f4132d7f84efc8cdb7f2759.png)
Thus the answer is
.
![$n=1, \lfloor 100n \rfloor = 100$](http://latex.artofproblemsolving.com/c/5/9/c598f41efc51873c90903f09dd03ade69d5279b2.png)