Find the value of: $i^{2} + i^{4} + i^{6}$ +..... upto $(2 n + 1)$ terms.

i

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1

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Solution

The correct option is D $- 1$
$i^{2}$ + $i^{4}$ + $i^{6}$ + ...upto $(2 n + 1)$ terms

$= (- 1 + 1) + (- 1 + 1) + (- 1 +$ .....upto $(2 n + 1))$ terms $\Rightarrow$ (odd terms)

$= 0 + 0 + . . . . - 1$

$= - 1$

Hence, 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

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

i^{2} = - 1

1 + i^{2} + i^{4} + i^{6} + i^{8} + . . .

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

= ?

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