+0  
 
0
178
6
avatar+644 

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

waffles  Oct 28, 2017
Sort: 

6+0 Answers

 #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

Guest Oct 28, 2017
 #2
avatar+92470 
+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

 

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

 

 

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

 

 

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

 

 

 

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

Melody  Oct 28, 2017
edited by Melody  Oct 28, 2017
 #6
avatar+92470 
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

Guest Oct 28, 2017
 #4
avatar+92470 
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  ?????.

Guest Oct 28, 2017

12 Online Users

avatar
avatar

New Privacy Policy (May 2018)

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see cookie policy and privacy policy.