If $i^{2} = - 1$ , then $1 + i^{2} + i^{4} + i^{6} + i^{8} + . . .$ to (2n+1) terms :

0

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+ 1

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3 i

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4 i

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Solution

The correct option is D $+ 1$
$S_{2 n} = i^{2} + i^{4} + i^{6} + i^{8} + . . . (i^{2})^{2 n + 1}$
The above series is a G.P with a common ratio of $i^{2}$ .
Hence
$S_{2 n} = \frac{(i^{2})^{2 n + 1} - 1}{i^{2} - 1}$
$= \frac{i^{4 n + 2} - 1}{i^{2} - 1}$
$= \frac{i^{2} (i^{4})^{n} - 1}{- 1 - 1}$
$= \frac{i^{2} - 1}{- 2}$
$= \frac{- 2}{- 2}$
$= 1$

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