$1 + i^{2} + i^{4} + i^{6} + . . . . . + i^{2 n}$ is [EAMCET 1980]

Positive

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Negative

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Zero

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Cannot be determined

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Solution

The correct option is D Cannot be determined
$S = 1 + i^{2} + i^{4} + . . . . . i^{2 n} = 1 - 1 + 1 - 1 + . . . . + (- 1)^{n}$
Obviously it depends on n.
Hence cannot be determined unless n is known.

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