Let $n$ be a positive integer. Then ${(i)}^{4 n + 1} + {(- i)}^{4 n + 5} =$

0

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2 i

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i

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- i

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Solution

The correct option is C $0$
$i^{4 n + 1} + (- i)^{4 n + 5}$
$= i (i^{4 n}) + (- i)^{5} (i^{4 n})$
$= i (i^{4})^{n} - i (i^{4})^{n}$
$= i (1)^{n} - i (1)^{n}$
$= i - i$
$= 0$ .

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