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The value of $\sum_{n=0}^{\infty}\left(\frac{2 i}{3}\right)^n$ is
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The correct answer is:
$\frac{9+6 i}{13}$
$\begin{aligned} & \sum_{n=0}^x\left(\frac{2 i}{3}\right)^n \\ &=1+\left(\frac{2 i}{3}\right)+\left(\frac{2 i}{3}\right)^2+\left(\frac{2 i}{3}\right)^3+\ldots . \\ &=\frac{1}{1-\frac{2 i}{3}}=\frac{3}{3-2 i} \times \frac{3+2 i}{3+2 i}\end{aligned}$
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