QUESTION

# If $\omega \ne 1$ is a cube root of unity then find the sum of the series $S = 1 + 2\omega + 3{\omega ^2} + ... + 3n{\omega ^{3n - 1}}$.A. $\dfrac{{3n}}{{\omega - 1}}$B. $3n(\omega - 1)$C. $\dfrac{{\omega - 1}}{{3n}}$D. 0

Hint: To solve this problem we need to know the properties of the cube root of 1, that is the sum of its roots are zero.

The given sum of series is $S = 1 + 2\omega + 3{\omega ^2} + ... + 3n{\omega ^{3n - 1}}$ ... (1)
Multiplying the above equation with $\omega$,
$\Rightarrow S\omega = \omega + 2{\omega ^2} + .... + (3n - 1){\omega ^{3n - 1}} + 3n{\omega ^{3n}}$ .... (2)
Taking the difference equation (1) – (2)
$\Rightarrow S - S\omega = 1 + \omega + {\omega ^2} + .... + {\omega ^{3n - 1}} - 3n{\omega ^{3n}} \to (3)$
From the properties of the cube root of 1, we have $1 + \omega + {\omega ^2}$=0 and ${\omega ^3} = 1$.
$\left[ {\because {\omega ^3} = 1 \Rightarrow {\omega ^{3n}} = 1} \right]$
Applying the properties on equation 3,
Since, all the terms till ${\omega ^{3n - 1}}$become 0. Hence we will be left with
$\Rightarrow S(1 - \omega ) = 0 - 3n$
Simplifying further,
$\Rightarrow S = \dfrac{{ - 3n}}{{1 - \omega }} = \dfrac{{3n}}{{\omega - 1}}$
$\therefore$ The sum of the series is $S = \dfrac{{3n}}{{\omega - 1}}$.
So, option A is the required option.

Note: The root of unity is a number which raised to the power of 3 gives the result as 1. According to the properties of the cube root of 1, the sum of its roots are zero. So $1 + \omega + {\omega ^2}$=0 and ${\omega ^3} = 0$ are used to solve the given equation to find the sum of series.