The value of sum_{n=1}^{13}(i^{n}+i^{n+1}), w | vedclass

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(B) We have $\sum_{n=1}^{13}(i^{n}+i^{n+1}) = \sum_{n=1}^{13} i^{n} + \sum_{n=1}^{13} i^{n+1}$.Since $i^{n}$ is a geometric progression with first ter

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$i-1$

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(B) We have $\sum_{n=1}^{13}(i^{n}+i^{n+1}) = \sum_{n=1}^{13} i^{n} + \sum_{n=1}^{13} i^{n+1}$.Since $i^{n}$ is a geometric progression with first term $a=i$ and common ratio $r=i$, the sum of $13$ terms is $S_{13} = i \frac{1-i^{13}}{1-i}$.Note that $i^{13} = (i^{4})^{3} 	imes i = 1^{3} 	imes i = i$.Thus, $\sum_{n=1}^{13} i^{n} = i \frac{1-i}{1-i} = i$.Similarly, $\sum_{n=1}^{13} i^{n+1} = i^{2} \frac{1-i^{13}}{1-i} = -1 \frac{1-i}{1-i} = -1$.Therefore, the total sum is $i + (-1) = i-1$.

The value of $\sum_{n=1}^{13}(i^{n}+i^{n+1})$, where $i=\sqrt{-1}$, is

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