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

Latest revision as of 09:51, 4 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 $\lfloor 100n \rfloor = \boxed{100}$ .

@@ Line 1: / Line 1: @@
-Given that <math>\displaystyle  iz^2=1+\frac 2z + \frac{3}{z^2}+\frac{4}{z ^3}+\frac{5}{z^4}+\cdots</math> and <math>\displaystyle z=a\pm \sqrt{b+i},</math> find <math>\displaystyle  \lfloor 100ab \rfloor.</math>
+==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>.
+==Solution==
+Multiplying both sides of the equation by <math>z</math>, we get
+<center><math>iz^3 = z + 2 + \frac{3}{z} + \frac{4}{z^2} + \cdots</math>,</center>
+and subtracting the original equation from this one we get
+<center><math>iz^2(z-1)=z+1+\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}+\cdots</math>.</center>
+Using the formula for an infinite geometric series, we find
+<center><math>iz^2(z-1)=\frac{z}{1-\frac{1}{z}}=\frac{z^2}{z-1}</math>.</center>
+Rearranging, we get
+<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>
+Thus <math>n=1</math>, and the answer is <math>\lfloor 100n \rfloor = \boxed{100}</math>.
+==See also==
+{{Mock AIME box|year=2006-2007|n=2|num-b=4|num-a=6}}