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Q.

$\sum_{n = 0}^{\infty} {(\frac{2 i}{3})}^{n} =$

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a

$9 + 6 i$

b

$9 - 6 i$

c

$\frac{9 + 6 i}{13}$

d

$\frac{9 - 6 i}{13}$

answer is A.

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Detailed Solution

$\begin{array}{rcl} \sum_{n = 0}^{\infty} {(\frac{2 i}{3})}^{n} & = & 1 + \frac{2 i}{3} + {(\frac{2 i}{3})}^{2} + {(\frac{2 i}{3})}^{3} + . . . . . . . . \\ = & \frac{1}{1 - (\frac{2 i}{3})} \\ = & \frac{3}{3 - 2 i} = \frac{3 (3 + 2 i)}{(3 - 2 i) (3 + 2 i)} \\ = & \frac{3 (3 + 2 i)}{13} \end{array}$

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