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