If omega neq 1 is a cube root of unity, then | vedclass

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(A) Given the series $S = 1 + 2\omega + 3\omega^2 + \dots + 3n\omega^{3n-1}$.Multiply by $\omega$: $S\omega = \omega + 2\omega^2 + 3\omega^3 + \dots +

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

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(A) Given the series $S = 1 + 2\omega + 3\omega^2 + \dots + 3n\omega^{3n-1}$.Multiply by $\omega$: $S\omega = \omega + 2\omega^2 + 3\omega^3 + \dots + 3n\omega^{3n}$.Subtracting the two equations: $S(1 - \omega) = 1 + \omega + \omega^2 + \dots + \omega^{3n-1} - 3n\omega^{3n}$.Since $\omega^3 = 1$, the sum of the geometric series $1 + \omega + \dots + \omega^{3n-1}$ is $\frac{1 - (\omega^3)^n}{1 - \omega} = \frac{1 - 1}{1 - \omega} = 0$.Thus, $S(1 - \omega) = 0 - 3n(1) = -3n$.Therefore, $S = \frac{-3n}{1 - \omega} = \frac{3n}{\omega - 1}$.

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

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