Question

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

• A. $-1$
• B. $1$
• C. $-i$
• D. $i$

Answer 1

The correct option is A $-1$

Consider $\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\frac{27}{128}+...$
which can be written as
$\frac{3^0}{2^1}+\frac{3^1}{2^3}+\frac{3^2}{2^5}+\frac{3^3}{2^7}+.......=\frac{1}{2}[1+\frac{3^{}}{2^2}+\frac{3^2}{2^4}+\frac{3^3}{2^6}+.......]$
Since $[1+\frac{3^{}}{2^2}+\frac{3^2}{2^4}+\frac{3^3}{2^6}+.......]$ 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
$\omega +{\omega }^{(\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\frac{27}{128}+....)}=\omega +{\omega }^2=-1[\because 1+\omega +{\omega }^2=0]$