Question

If $\omega \ne 1$ is a cube root of unity, then the sum of the series $S=1+2\omega +3{\omega }^2+..........+3n{\omega }^{3n-1}$ is

• A. $\frac{3n}{\omega -1}$
• B. $3n\left(\omega -1\right)$
• C. $\frac{\left(\omega -1\right)}{3n}$
• D. $0$

Answer 1

Answer: (A)

Given, $S=1+2\omega +3{\omega }^2+\dots +3n{\omega }^{3n-1}$
$\therefore S\omega =\omega +2{\omega }^2+\dots +(3n-1){\omega }^{3n}+3n{\omega }^{3n}$
$\Rightarrow S(1-\omega )=1+\omega +{\omega }^2+\dots +{\omega }^{3n-1}+3n{\omega }^{3n}$
$\Rightarrow S(1-\omega )=0-3n$
$\Rightarrow S=-\frac{3n}{1-\omega }=\frac{3n}{\omega -1}$