Question

If $\omega$ is the complex cube root of unity, then the value of $\omega +\omega \left(\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\frac{27}{128}+\dots \dots \right)$

• A. -1
• B. 1
• C. -i
• D. i

Answer 1

Answer: (A)

Consider $\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\frac{27}{128}+\dots$
which can be written as
$\frac{3^0}{2^1}+\frac{3^1}{2^3}+\frac{3^2}{2^5}+\frac{3^3}{2^7}+...$
$=\frac{1}{2}\left[1+\frac{3}{2^2}+\frac{3^2}{2^4}+\frac{3^3}{2^6}+...\right]$
Since $\left[1+\frac{3}{2^2}+\frac{3^2}{2^4}+\frac{3^3}{2^6}+\dots \dots \right]$ is a GP therefore by sum of infinite GP, we have
$=\frac{1}{2}\left[\frac{1}{1-\frac{3}{2^2}}\right]=2$
Therefore, given expression is
$\omega +\omega \left(\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\frac{27}{128}+...\right)$
$=\omega +{\omega }^2=-1$
$\left[\because 1+\omega +{\omega }^2=0\right]$