Question:
Evaluate
$\sum_{n=1}^{13}\left(i^{n}+i^{n+1}\right)$, where $n \in N$
Solution:
According to the question,
We have,
$\sum_{n=1}^{13}\left(i^{n}+i^{n+1}\right)=\sum_{n=1}^{13}(1+i) i^{n}$
$=(1+i)\left(1+i^{2}+i^{3}+i^{4}+i^{5}+i^{6}+i^{7}+i^{8}+i^{9}+i^{10}+i^{11}+i^{12}+i^{13}\right)$
$=(1+i) \frac{i\left(i^{13}-1\right)}{i-1}$
$=(1+i) \frac{i(i-1)}{i-1}$
$=(1+i) i$
$=i+i^{2}$
$=i-1$
$\therefore \sum_{n=1}^{13}\left(i^{n}+i^{n+1}\right)=1-1$
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