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

Question

If $\omega \neq 1$ is a cube root of unity, then the sum of the series $S=1+2 \omega+3 \omega^2+\ldots \ldots \ldots . .+3 n \omega^{3 n-1}$ is, $\frac{3 n}{\omega-1}$, $3 n(\omega-1)$, $\frac{\omega-1}{3 n}$, 0

MARKS App · Accepted Answer

$\begin{aligned} \text {Hints : } & s=1+2 \omega+3 \omega^2+\ldots \ldots \ldots .+3 n \omega^{3 n-1} \ & s \omega=\omega+2 \omega^2+\ldots \ldots \ldots \ldots .+(3 n-1) \omega^{3 n}+3 n \omega^{3 n} \ & s(1-\omega)=1+\omega+\omega^2+\ldots \ldots \ldots .+\omega^{3 n-1}-3 n \omega^{3 \mathrm{~h}} \ & =0-3 n \ & s=\frac{-3 n}{1-\omega}=\frac{3 n}{\omega-1} \end{aligned}$

Search any question & find its solution

Looking for more such questions to practice?