Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 5"
m
 
(3 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 
==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> 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==
 
==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
 +
<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==
 
==See also==
−
*[[Mock AIME 2 2006-2007/Problem 4 | Previous Problem]]
+
{{Mock AIME box|year=2006-2007|n=2|num-b=4|num-a=6}}
−
*[[Mock AIME 2 2006-2007/Problem 6 | Next Problem]]
 
−
*[[Mock AIME 2 2006-2007]]
 

Latest revision as of 09:51, 4 April 2012

Problem

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 $\lfloor 100n \rfloor$.

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 $n=1$, and the answer is $\lfloor 100n \rfloor = \boxed{100}$.

See also

Mock AIME 2 2006-2007 (Problems, Source)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_2_2006-2007_Problems/Problem_5&oldid=46094"