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Question Number 185165 by aba last updated on 17/Jan/23

$$\mathrm{Simplify}:\frac{{\mathrm{i}}^{2022}+{\mathrm{i}}^{2023}+{\mathrm{i}}^{2024}+{\mathrm{i}}^{2025}}{1+\mathrm{i}}$$

Answered by HeferH last updated on 18/Jan/23

$$if:i=\sqrt{-1}$$$$i^2=-1$$$$i^3=-i$$$$\Rightarrow$$$$\frac{i^{2022}(1+i+i^2+i^3)}{1+i}=\frac{i^{2022}(1+i-1-i)}{1+i}=0$$

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