Question

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

• (A) 3n/ω - 1
• (B) 3n (ω - 1)
• (C) ω-1/3n
• (D) 0

Answer 1

The Correct Option is A

Explanation:
$$\begin{array}{c}s=1+2\omega +3{\omega }^2+\dots \dots \dots +3n{\omega }^{3n-1}\\ s\omega =\omega +2{\omega }^2+\dots \dots \dots \dots +(3n-1){\omega }^{3n}+3n{\omega }^{3n}\\ s(1-\omega )=1+\omega +{\omega }^2+\dots \dots \dots +{\omega }^{3n-1}-3n{\omega }^{3k}\\ =0-3n\\ s=\frac{-3n}{1-\omega }=\frac{3n}{\omega -1}\end{array}$$