+0  
 
0
341
6
avatar+644 

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

 Oct 28, 2017
 #1
avatar
0

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

 Oct 28, 2017
 #2
avatar+96956 
+2

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

 

This is the sume of a  GP

a=1, r=i, n=2010

 

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

 

 

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

 

 

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

 

 

 

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

.
 Oct 28, 2017
edited by Melody  Oct 28, 2017
 #6
avatar+96956 
0

PLEASE NOTE - I HAVE CORRECTED A CARELESS ERROR 

 

Thank you Alan for alerting me to it   laugh

Melody  Oct 28, 2017
 #3
avatar
0

Melody: Can you please explain this "goofy" answer by W/A? Thanks.

 

∑[1 + i^n], n=1 to 2009 =2,009 + i

 Oct 28, 2017
 #4
avatar+96956 
0

Nope I have no idea :)

Melody  Oct 28, 2017
 #5
avatar
0

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  ?????.

 Oct 28, 2017

17 Online Users

avatar
avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.