sumlimits_{n = 1}^{50} {{i^{2n-1}}} is equal | vedclass

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(A) The given sum is $S = \sum\limits_{n = 1}^{50} {{i^{2n - 1}}} = i^1 + i^3 + i^5 + i^7 + \dots + i^{99}$.This is a geometric progression with first

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(A) The given sum is $S = \sum\limits_{n = 1}^{50} {{i^{2n - 1}}} = i^1 + i^3 + i^5 + i^7 + \dots + i^{99}$.This is a geometric progression with first term $a = i$,common ratio $r = i^2 = -1$,and number of terms $n = 50$.The sum of a geometric progression is given by $S = \frac{a(1 - r^n)}{1 - r}$.Substituting the values: $S = \frac{i(1 - (-1)^{50})}{1 - (-1)} = \frac{i(1 - 1)}{2} = \frac{i(0)}{2} = 0$.

$\sum\limits_{n = 1}^{50} {{i^{2n-1}}}$ is equal to (where $i = \sqrt{-1}$)

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