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The value of $\sum_{n=1}^{13}\left(i^{n}+i^{n+1}\right), i=\sqrt{-1}$ is
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Verified Answer
The correct answer is:
$i-1$
We have.
$\begin{aligned} & \sum_{n=1}^{13}\left(i^{n}+i^{n+1}\right) \\=& \sum_{n=1}^{13} i^{n}+\sum_{n=1}^{13} i^{n+1} \\=& i \left(\frac{1-i^{13}}{1-i}\right)+i^{2}\left(\frac{1-i^{13}}{1-i}\right) \\=& i \left(\frac{1-i}{1-i}\right)-\left(\frac{1-i}{1-i}\right) \\=& i-1 \end{aligned}$
$\begin{aligned} & \sum_{n=1}^{13}\left(i^{n}+i^{n+1}\right) \\=& \sum_{n=1}^{13} i^{n}+\sum_{n=1}^{13} i^{n+1} \\=& i \left(\frac{1-i^{13}}{1-i}\right)+i^{2}\left(\frac{1-i^{13}}{1-i}\right) \\=& i \left(\frac{1-i}{1-i}\right)-\left(\frac{1-i}{1-i}\right) \\=& i-1 \end{aligned}$
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