$i^{2} + i^{4} + i^{6} + . . .$ upto $(2 n + 1)$ terms =

i

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- i

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1

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- 1

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Solution

The correct option is D $- 1$
$i^{2} + i^{4} + i^{6} + . . . . . . . . i^{2 n + 1}$
$- 1 + 1 - 1 + . . . . (i^{2 n x})$
odd no : of terms
$\Rightarrow - 1$
$\Rightarrow - 1 + 1 - 1 + 1 . . . . + 1 - 1$
$\Rightarrow - 1$
option $D$ is correct.

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Similar questions

i^{2} + i^{4} + i^{6}

(2 n + 1)

$i^{2}$ + $i^{4}$ + $i^{6}$ + ........ upto (2n+1) terms =

i^{2} = - 1

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i^{2} + i^{4} + i^{6} + . . . + (2 n + 1)

i^{2} = - 1

i^{2} + i^{4} + i^{6} + i^{8} + . . . + (2 n)

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