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. cookie policy and privacy policy.
 
+0  
 
0
247
2
avatar

How do you solve

$i^{-100}+i^{-99}+i^{-98}+\cdots+i^{-1}+i^0+i^1+\cdots+i^{99}+i^{100}$

 Jun 11, 2018
 #1
avatar+985 
+3

\(i^{-100}+i^{-99}+i^{-98}+\cdots+i^{-1}+i^0+i^1+\cdots+i^{99}+i^{100}\)

 

\(i=\sqrt{-1}, \text{ we start simple and try to find a pattern.}\)

 

\(i^0=1\\ i^1=i\\ i^2=-1\\ i^3=-i\\ i^4=1\\ i^5=i\\ i^6=-1\\ i^7=-i\\ \)

Every 4 terms, the sequence rotates, \(1, i , -1, -i\)

 

Their sum is: \(1+i-1-i=0\)

 

Since every four terms cancel out, we just need to know what term the sequence starts and ends. 

 

\(i^{-100}=1\\ i^{100}=1\\\)

\(i^{-100}+i^{-99}+i^{-98}+\cdots+i^{-1}+i^0+i^1+\cdots+i^{99}+i^{100}\\ =1+i-1-i+1+-\cdots+1+i-1-i+1\\ =\boxed1\)

 

I hope this helped,

 

Gavin

 Jun 11, 2018
edited by GYanggg  Jun 11, 2018
 #2
avatar+22343 
+1

How do you solve
$i^{-100}+i^{-99}+i^{-98}+\cdots+i^{-1}+i^0+i^1+\cdots+i^{99}+i^{100}$

 

\(\small{ \begin{array}{|rcll|} \hline && i^{-100}+i^{-99}+i^{-98}+i^{-97}+\cdots+i^{-2}+i^{-1}+i^0+i^1+i^2+i^3+i^4\cdots+i^{99}+i^{100} \\ &=& i^{-100}(1+i)+i^{-98}(1+i) +\cdots+i^{-2}(1+i)+i^0+i^1(1+i)+i^3(1+i)+\cdots+i^{99}(1+i) \\ && \boxed{\color{red}1+i =1-1=0} \\ &=& i^{-100}\times 0+i^{-98}\times 0 +\cdots+i^{-2}\times 0+i^0+i^1\times 0+i^3\times 0+\cdots+i^{99}\times 0 \\ &=&0+i^0+0 \\ &=& i^0 \\ &\mathbf{=}& \mathbf{1} \\ \hline \end{array} }\)

 

laugh

 Jun 12, 2018
edited by heureka  Jun 12, 2018

18 Online Users

avatar
avatar
avatar
avatar