How do you solve
$i^{-100}+i^{-99}+i^{-98}+\cdots+i^{-1}+i^0+i^1+\cdots+i^{99}+i^{100}$
+981 \(i^{-100}+i^{-99}+i^{-98}+\cdots+i^{-1}+i^0+i^1+\cdots+i^{99}+i^{100}\)
\(i=\sqrt{-1}, \text{ we start simple and try to find a pattern.}\)
\(i^0=1\\ i^1=i\\ i^2=-1\\ i^3=-i\\ i^4=1\\ i^5=i\\ i^6=-1\\ i^7=-i\\ \)
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
+26367 How do you solve
$i^{-100}+i^{-99}+i^{-98}+\cdots+i^{-1}+i^0+i^1+\cdots+i^{99}+i^{100}$
\(\small{ \begin{array}{|rcll|} \hline && i^{-100}+i^{-99}+i^{-98}+i^{-97}+\cdots+i^{-2}+i^{-1}+i^0+i^1+i^2+i^3+i^4\cdots+i^{99}+i^{100} \\ &=& i^{-100}(1+i)+i^{-98}(1+i) +\cdots+i^{-2}(1+i)+i^0+i^1(1+i)+i^3(1+i)+\cdots+i^{99}(1+i) \\ && \boxed{\color{red}1+i =1-1=0} \\ &=& i^{-100}\times 0+i^{-98}\times 0 +\cdots+i^{-2}\times 0+i^0+i^1\times 0+i^3\times 0+\cdots+i^{99}\times 0 \\ &=&0+i^0+0 \\ &=& i^0 \\ &\mathbf{=}& \mathbf{1} \\ \hline \end{array} }\)
