Q155 / 270 · NDA 2016 (Phase 2) · Mathematics

What is $\omega^{100}+\omega^{200}+\omega^{300}$ equal to, where $\omega$ is the cube root of unity?

  • A
    $1$
  • B
    $3 \omega$
  • C
    $3 \omega^2$
  • Correct answer
    D
    $0$

Solution

$\begin{aligned} & \omega^{100}+\omega^{200}+\omega^{300}=\left(\omega^{99} \cdot \omega\right)+\left(\omega^{100}\right)^2+\left(\omega^3\right)^{100} \\ & =\left(\omega^{99} \cdot \omega\right)+\left(\omega^{99} \cdot \omega\right)^2+\left(\omega^3\right)^{100} \\ & \because \omega^3=1 \quad \omega^{99}=\left(\omega^3\right)^{33}=1^{33}=1 \\ & \Rightarrow \omega^{100}+\omega^{200}+\omega^{300}=(1 \cdot \omega)+(1 \cdot \omega)^2+1^{100} \\ & \quad=\omega+\omega^2+1=1+\omega+\omega^2=0\end{aligned}$

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

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