Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 5"
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==Problem== | ==Problem== | ||
− | + | Given that <math> iz^2=1+\frac 2z + \frac{3}{z^2}+\frac{4}{z ^3}+\frac{5}{z^4}+\cdots</math> and <math>z=n\pm \sqrt{-i},</math> find <math> \lfloor 100n \rfloor.</math> | |
− | Given that <math> | ||
==Solution== | ==Solution== | ||
− | |||
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> | 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> | ||
− | + | ==See also== | |
− | |||
*[[Mock AIME 2 2006-2007/Problem 4 | Previous Problem]] | *[[Mock AIME 2 2006-2007/Problem 4 | Previous Problem]] | ||
− | |||
*[[Mock AIME 2 2006-2007/Problem 6 | Next Problem]] | *[[Mock AIME 2 2006-2007/Problem 6 | Next Problem]] | ||
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*[[Mock AIME 2 2006-2007]] | *[[Mock AIME 2 2006-2007]] |
Revision as of 01:33, 23 April 2008
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 the answer is
.
