If i=sqrt{-1},then sum_{n=0}^{infty}left(frac | vedclass

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(D) The given expression is an infinite geometric series: $\sum_{n=0}^{\infty}\left(\frac{i}{3}\right)^n = 1 + \frac{i}{3} + \left(\frac{i}{3}\right)^

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$\frac{9+3i}{10}$

Vedclass · Answer

(D) The given expression is an infinite geometric series: $\sum_{n=0}^{\infty}\left(\frac{i}{3}ight)^n = 1 + \frac{i}{3} + \left(\frac{i}{3}ight)^2 + \dots \infty$. The sum of an infinite geometric series is given by $S = \frac{a}{1-r}$,where $a=1$ and $r=\frac{i}{3}$. $S = \frac{1}{1-\frac{i}{3}} = \frac{1}{\frac{3-i}{3}} = \frac{3}{3-i}$. To simplify,multiply the numerator and denominator by the conjugate $(3+i)$: $S = \frac{3}{3-i} 	imes \frac{3+i}{3+i} = \frac{3(3+i)}{3^2 - i^2} = \frac{9+3i}{9 - (-1)} = \frac{9+3i}{10}$.

If $i=\sqrt{-1}$,then $\sum_{n=0}^{\infty}\left(\frac{i}{3}\right)^n=$

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