Question

$\omega +{\omega }^{\left(\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\frac{27}{128}+........\right)}$ is equal to

• A. $-1$
• B. $0$
• C. $9$
• D. $-i$

Answer 1

The correct option is A $-1$

Let us take $\left(\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\frac{27}{128}+........\right)$
Above series is an infinite geometric series.
$\Rightarrow S_{\infty }=\left(\frac{1/2}{1-3/4}\right)=2$
$\therefore \omega +{\omega }^{\left(\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\frac{27}{128}+........\right)}$
$=\omega +{\omega }^2=-1$
Hence, option A.