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Simplify \(i^1+i^2+i^3+\cdots+ i^{97} + i^{98}+i^{99}.\)

 Sep 22, 2018

Best Answer 

 #1
avatar+5226 
+2

\(i^{4k+2}+i^{4k+4}=0 \\ i^{4k+1}+i^{4k+3}=0\)

 

\(\text{we can rewrite the sum as}\\ (i + i^3) + (i^2 + i^4) + (i^5 + i^7) + (i^6 + i^8) + \dots + (i^{98}+i^{100}) +(i^{97}+ i^{99}) - i^{100} = -i^{100} \\ -i^{100} =- i^{4(25)} = -1\)

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 Sep 22, 2018
 #1
avatar+5226 
+2
Best Answer

\(i^{4k+2}+i^{4k+4}=0 \\ i^{4k+1}+i^{4k+3}=0\)

 

\(\text{we can rewrite the sum as}\\ (i + i^3) + (i^2 + i^4) + (i^5 + i^7) + (i^6 + i^8) + \dots + (i^{98}+i^{100}) +(i^{97}+ i^{99}) - i^{100} = -i^{100} \\ -i^{100} =- i^{4(25)} = -1\)

Rom Sep 22, 2018

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