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Compute 1+i+i^2+i^3+i^4+....i^2009

waffles  Oct 28, 2017
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 #1
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Everything cancels out except 2009 positive "1s" PLUS the last term of: i^2009 =i. Therefore the answer is: =2,009 + i

Guest Oct 28, 2017
 #2
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Compute 1+i+i^2+i^3+i^4+....i^2009

 

This is the sume of a  GP

a=1, r=i, n=2010

 

\(\boxed{s_n=\frac{a(1-r^n)}{1-r}}\\ S_{2010}=\frac{1(1-i^{2010})}{1-i}\\ S_{2010}=\frac{1-i^{2010}}{1-i}\\\)

 

 

\(i^1=i \\ i^2=-1\\ i^3=-i\\ i^4=+1\\ ...For \;\;k\in Z \\i^{4k+1}=i\\ i^{4k+2}=-1\\ i^{4k+3}=-i\\ i^{4k}=1\\ so\\ i^{2010}=i^{4*{502}+2}=-1\)

 

 

\(S_{2010}=\frac{1-i^{2010}}{1-i}\\ S_{2010}=\frac{1-\color{red}{-1}}{1-i}\qquad \text {Error corrected here}\\ S_{2010}=\frac{2}{1-i}\\ S_{2010}=\frac{2}{1-i}\times\frac{1+i}{1+i} \\ S_{2010}=\frac{2(1+i)}{1-i^2}\\ S_{2010}=\frac{2(1+i)}{1--1}\\ S_{2010}=\frac{2(1+i)}{2}\\ S_{2010}=1+i \)

 

 

 

\(1+i+i^2+i^3+i^4+....i^{2009}=1+i\)

Melody  Oct 28, 2017
edited by Melody  Oct 28, 2017
 #6
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PLEASE NOTE - I HAVE CORRECTED A CARELESS ERROR 

 

Thank you Alan for alerting me to it   laugh

Melody  Oct 28, 2017
 #3
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Melody: Can you please explain this "goofy" answer by W/A? Thanks.

 

∑[1 + i^n], n=1 to 2009 =2,009 + i

Guest Oct 28, 2017
 #4
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Nope I have no idea :)

Melody  Oct 28, 2017
 #5
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OK. I think I found the mistake !! The first 1 should be outside the "Sigma" sign, but it still gives a different answer: 1 + ∑{ i^n}, n=1 to 2009 = 1 + i  ?????.

Guest Oct 28, 2017

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