Question

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

• A. -1
• B. 1
• C. -i
• D. i

Answer 1

Answer: (A)

Consider $\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\frac{27}{128}+\ldots$
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 $\left[1+\frac{3}{2^{2}}+\frac{3^{2}}{2^{4}}+\frac{3^{3}}{2^{6}}+\ldots \ldots\right]$ 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
$\omega+\omega\left(\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\frac{27}{128}+...\right)$
$=\omega+\omega^{2}=-1$
$\left[\because 1+\omega+\omega^{2}=0\right]$