sumlimits_{n=1}^{50} i^{(2n-1)!} is equal to | vedclass

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(C) The given sum is $S = \sum_{n=1}^{50} i^{(2n-1)!} = i^{1!} + i^{3!} + i^{5!} + i^{7!} + \dots + i^{99!}$.For $n=1$,$i^{1!} = i^1 = i$.For $n=2$,$i

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$47 + i$

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(C) The given sum is $S = \sum_{n=1}^{50} i^{(2n-1)!} = i^{1!} + i^{3!} + i^{5!} + i^{7!} + \dots + i^{99!}$.For $n=1$,$i^{1!} = i^1 = i$.For $n=2$,$i^{3!} = i^6 = i^4 	imes i^2 = 1 	imes (-1) = -1$.For $n \ge 3$,$(2n-1)!$ is a multiple of $4$ (since $(2n-1)!$ contains $4 	imes 3 	imes 2 	imes 1$ as factors for $n \ge 3$).Thus,$i^{(2n-1)!} = (i^4)^k = 1^k = 1$ for all $n \ge 3$.There are $50 - 3 + 1 = 48$ terms from $n=3$ to $n=50$,each equal to $1$.Therefore,$S = i + (-1) + 48(1) = i - 1 + 48 = 47 + i$.

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

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