If ω ≠ 1 is a cube root of unity, then the sum of the series S=1+2 ω+3 ω^2+… … … . .+3 n ω^3 n-1 is

Correct answer: (A) (3 n)/(ω-1). Hints : s=1+2 ω+3 ω^2+… … … .+3 n ω^3 n-1 s ω=ω+2 ω^2+… … … … .+(3 n-1) ω^3 n+3 n ω^3 n s(1-ω)=1+ω+ω^2+… … … .+ω^3 n-1-3 n ω^3 ~h =0-3 n s=(-3 n)/(1-ω)=(3 n)/(ω-1)

Q96 / 160 · WBJEE 2011 · Mathematics

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

Correct answer
A
\(\frac{3 n}{\omega-1}\)
B
\(3 n(\omega-1)\)
C
\(\frac{\omega-1}{3 n}\)
D
0

Solution

\(\begin{aligned} \text {Hints : } & s=1+2 \omega+3 \omega^2+\ldots \ldots \ldots .+3 n \omega^{3 n-1} \\ & s \omega=\omega+2 \omega^2+\ldots \ldots \ldots \ldots .+(3 n-1) \omega^{3 n}+3 n \omega^{3 n} \\ & s(1-\omega)=1+\omega+\omega^2+\ldots \ldots \ldots .+\omega^{3 n-1}-3 n \omega^{3 \mathrm{~h}} \\ & =0-3 n \\ & s=\frac{-3 n}{1-\omega}=\frac{3 n}{\omega-1} \end{aligned}\)

Hence, the correct answer is option: 'A'.

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

Solution

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