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If $i=\sqrt{-1}$ then $1+\mathrm{i}^2+\mathrm{i}^4+\mathrm{i}^6+\ldots+{ }^1 \mathrm{i}^{2024}=$
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$1$
$1+i^2+i^4+i^6+\ldots \ldots+i^{2024}$
$\begin{aligned} & i^{4 n}=1 \\ & i^{2 n}=-1 \\ & \left(1+i^2\right)+\left(i^4+i^6\right)+\ldots .+\left(i^{2020}+i^{2022}\right)+i^{2024} \\ & =0+0+\ldots . .+0+1=0+1=1\end{aligned}$
$\begin{aligned} & i^{4 n}=1 \\ & i^{2 n}=-1 \\ & \left(1+i^2\right)+\left(i^4+i^6\right)+\ldots .+\left(i^{2020}+i^{2022}\right)+i^{2024} \\ & =0+0+\ldots . .+0+1=0+1=1\end{aligned}$
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