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




Question Number 185165 by aba last updated on 17/Jan/23
Simplify : ((i^(2022) +i^(2023) +i^(2024) +i^(2025) )/(1+i))
$$\mathrm{Simplify}\::\:\frac{\mathrm{i}^{\mathrm{2022}} +\mathrm{i}^{\mathrm{2023}} +\mathrm{i}^{\mathrm{2024}} +\mathrm{i}^{\mathrm{2025}} }{\mathrm{1}+\mathrm{i}} \\ $$
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
$$\:{if}\::\:{i}\:=\:\sqrt{−\mathrm{1}} \\ $$$$\:{i}^{\mathrm{2}} =\:−\mathrm{1} \\ $$$$\:{i}^{\mathrm{3}} =\:−{i}\: \\ $$$$\:\:\Rightarrow\: \\ $$$$\:\frac{{i}^{\mathrm{2022}} \left(\mathrm{1}+{i}+{i}^{\mathrm{2}} +{i}^{\mathrm{3}} \right)}{\mathrm{1}+{i}}\:=\:\frac{{i}^{\mathrm{2022}} \left(\mathrm{1}+{i}\:−\mathrm{1}\:−{i}\right)}{\mathrm{1}\:+\:{i}}=\:\mathrm{0} \\ $$

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