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

Answer 1

Answer: (A)

$\frac{3n}{\omega -1}$

Given $S=1+2\omega +3{\omega }^2+.....+3n{\omega }^{3n-1}$
$\therefore S\omega =\omega +2{\omega }^2+.....+(3n-1){\omega }^{3n-1}+3n{\omega }^{3n}$
$\Rightarrow S(1-\omega )=1+\omega +{\omega }^2+.....+{\omega }^{3n-1}-3n{\omega }^{3n}$
$\Rightarrow S(1-\omega )=0-3n$ ..... $[\because {\omega }^3=1⟹{\omega }^{3n}=1]$
$\Rightarrow S=-\frac{3n}{1-\omega }=\frac{3n}{\omega -1}$