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

Revision as of 12:43, 16 September 2006

Problem

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

Solution

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 the answer is $n=1, \lfloor 100n \rfloor = 100$ .

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Mock AIME 2 2006-2007

@@ Line 4: / Line 4: @@
 ==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>
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Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 5"

Revision as of 12:43, 16 September 2006

Problem

Solution