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

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Solution

The correct option is D $0$
$S_{2 n} = i^{2} + i^{4} + i^{6} + i^{8} + . . . (i^{2})^{2 n}$
The above series is a G.P with a common ratio of $i^{2}$ .
Hence
$S_{2 n} = \frac{(i^{2})^{2 n} - 1}{i^{2} - 1}$
$= \frac{i^{4 n} - 1}{i^{2} - 1}$
$= \frac{(i^{4})^{n} - 1}{- 1 - 1}$
$= \frac{1 - 1}{- 2}$
$= 0$

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