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 |
| − | + | <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}} | ||
Latest revision as of 09:51, 4 April 2012
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 
, and the answer is 
.
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 | ||
