WBJEE · Maths · Sequences and Series
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
- A \(\frac{3 n}{\omega-1}\)
- B \(3 n(\omega-1)\)
- C \(\frac{\omega-1}{3 n}\)
- D 0
Answer & Solution
Correct Answer
(A) \(\frac{3 n}{\omega-1}\)
Step-by-step Solution
Detailed explanation
\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…
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