The value of the sum sumlimits_{n = 1}^{13} { | vedclass

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(B) We are given the sum $S = \sum\limits_{n = 1}^{13} {({i^n} + {i^{n + 1}})} $.This can be written as $S = \sum\limits_{n = 1}^{13} {i^n} + \sum\lim

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$i - 1$

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(B) We are given the sum $S = \sum\limits_{n = 1}^{13} {({i^n} + {i^{n + 1}})} $.This can be written as $S = \sum\limits_{n = 1}^{13} {i^n} + \sum\limits_{n = 1}^{13} {i^{n + 1}}$.Both are geometric series with $13$ terms.For the first series,the sum is $\frac{i(1 - i^{13})}{1 - i}$. Since $i^4 = 1$,$i^{13} = i^{12} 	imes i = 1 	imes i = i$.So,the first sum is $\frac{i(1 - i)}{1 - i} = i$.For the second series,the sum is $\frac{i^2(1 - i^{13})}{1 - i} = \frac{-1(1 - i)}{1 - i} = -1$.Therefore,$S = i + (-1) = i - 1$.

The value of the sum $\sum\limits_{n = 1}^{13} {({i^n} + {i^{n + 1}})} $,where $i = \sqrt { - 1} $,equals

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