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Simplify-i-2022-i-2023-i-2024-i-2025-1-i-

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Question Number 185165 by aba last updated on 17/Jan/23
Simplify : ((i^(2022) +i^(2023) +i^(2024) +i^(2025) )/(1+i))
Simplify:i2022+i2023+i2024+i20251+i
Answered by HeferH last updated on 18/Jan/23
if : i = (√(−1)) i^2 = −1 i^3 = −i ⇒ ((i^(2022) (1+i+i^2 +i^3 ))/(1+i)) = ((i^(2022) (1+i −1 −i))/(1 + i))= 0
if:i=−1i2=−1i3=−i⇒i2022(1+i+i2+i3)1+i=i2022(1+i−1−i)1+i=0

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