Question

If omega $\left(\omega \right)\ne 1$ 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. $\frac{\omega -1}{3n}$
• D. $0$

Answer 1

The correct option is A

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

Explanation for the correct option.

Step 1: Prerequisites for the solution

Since $\omega$ is a cube root of unity, then we know that

${\omega }^3=1\text{and}1+\omega +{\omega }^2=0$

As the given series is

$S=1+2\omega +3{\omega }^2+...+3n{\omega }^{3n-1}$

To simplify multiply both the sides by $\omega$ then we have

$S\omega =\omega +2{\omega }^2+3{\omega }^3+...+3n{\omega }^{3n}$

Step 2: Calculation for the correct option

Now, we know the $S\text{and}S\omega$ then

$\begin{array}{rcl}S-S\omega & =& \left(1+2\omega +3{\omega }^2+\dots +3n{\omega }^{3n-1}\right)-\left(\omega +2{\omega }^2+3{\omega }^3+\dots +3n{\omega }^{3n}\right)\\ & =& 1+\omega +{\omega }^2+\dots +3n{\omega }^{3n-1}-3n{\omega }^{3n}\end{array}$

Here, the part $\begin{array}{rcl}& & 1+\omega +{\omega }^2+\dots +3n{\omega }^{3n-1}\end{array}$ is a Geometric Progression, then by the formula of the sum of a geometric progression we have,

$\begin{array}{rcl}1+\omega +{\omega }^2+\dots +3n{\omega }^{3n-1}& =& \frac{1\left({\omega }^{3n}-1\right)}{\omega -1}\end{array}$

From this, we can write that

$\begin{array}{rcl}S\left(1-\omega \right)& =& \frac{1\left({\omega }^{3n}-1\right)}{\omega -1}-3{\omega }^{3n}\end{array}$

Since ${\omega }^3=1\text{then}{\left({\omega }^3\right)}^n=1$ which implies that

$\begin{array}{rcl}S\left(1-\omega \right)& =& \frac{1\left(1-1\right)}{\omega -1}-3n\\ & =& -3n\\ & \Rightarrow & S=\frac{3n}{\omega -1}\end{array}$

Hence, the correct option is (A).