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If $\omega$ is a complex cube root of unity, then
$\begin{aligned} & \left(\frac{1-\sqrt{3} i}{2}\right)^{2020}+\left(\frac{1+\sqrt{3} i}{2}\right)^{2026} \\ & +\sin \left(\sum_{j=1}^6(j+\omega)\left(j+\omega^2\right) \frac{3 \pi}{152}\right)=\end{aligned}$
Options:
$\begin{aligned} & \left(\frac{1-\sqrt{3} i}{2}\right)^{2020}+\left(\frac{1+\sqrt{3} i}{2}\right)^{2026} \\ & +\sin \left(\sum_{j=1}^6(j+\omega)\left(j+\omega^2\right) \frac{3 \pi}{152}\right)=\end{aligned}$
Solution:
1517 Upvotes
Verified Answer
The correct answer is:
$-2$
We have,
$\frac{1-\sqrt{3} i}{2}=-\left[\frac{-1+\sqrt{3} i}{2}\right]=-\omega$
and $\frac{1+\sqrt{3} i}{2}=-\left[\frac{-1-\sqrt{3} i}{2}\right]=-\omega^2$
$\therefore \quad\left(\frac{1-\sqrt{3} i}{2}\right)^{2020}+\left(\frac{1+\sqrt{3} i}{2}\right)^{2026}$
$\begin{aligned} & =(-\omega)^{2020}+\left(-\omega^2\right)^{2026} \\ & =\omega^{2020}+\omega^{4052}=\omega+\omega^2=-1\end{aligned}$
Now,
$\begin{aligned} & \sum_{j=1}^6(j+\omega)\left(j+\omega^2\right) \\ & =\sum_{j=1}^6\left(j^2+j \omega^2+j \omega+\omega^3\right) \\ & =\sum_{j=1}^6\left(j^2-j+1\right)=\sum_{j=1}^6 j^2-\sum_{j=1}^6 j+\sum_{j=1}^6 1 \\ & =\frac{6(6+1)(12+1)}{6}-\frac{6(6+1)}{2}+6 \\ & =7 \times 13-3 \times 7+6=91-21+6=76\end{aligned}$
$\begin{aligned} \therefore \sin \left(\sum_{j=1}^6(j+\omega)\left(j+\omega^2\right) \frac{3 \pi}{152}\right) & =\sin \left(\frac{76 \times 3 \pi}{152}\right) \\ & =\sin \frac{3 \pi}{2}=-1\end{aligned}$
$\begin{aligned} \therefore \quad\left(\frac{1-\sqrt{3} i}{2}\right)^{2020}+ & \left(\frac{1+\sqrt{3} i}{2}\right)^{2026} \\ & +\sin \left(\sum_{j=1}^6(j+\omega)\left(j+\omega^2\right) \frac{3 \pi}{512}\right) \\ =-1-1 & =-2\end{aligned}$
$\frac{1-\sqrt{3} i}{2}=-\left[\frac{-1+\sqrt{3} i}{2}\right]=-\omega$
and $\frac{1+\sqrt{3} i}{2}=-\left[\frac{-1-\sqrt{3} i}{2}\right]=-\omega^2$
$\therefore \quad\left(\frac{1-\sqrt{3} i}{2}\right)^{2020}+\left(\frac{1+\sqrt{3} i}{2}\right)^{2026}$
$\begin{aligned} & =(-\omega)^{2020}+\left(-\omega^2\right)^{2026} \\ & =\omega^{2020}+\omega^{4052}=\omega+\omega^2=-1\end{aligned}$
Now,
$\begin{aligned} & \sum_{j=1}^6(j+\omega)\left(j+\omega^2\right) \\ & =\sum_{j=1}^6\left(j^2+j \omega^2+j \omega+\omega^3\right) \\ & =\sum_{j=1}^6\left(j^2-j+1\right)=\sum_{j=1}^6 j^2-\sum_{j=1}^6 j+\sum_{j=1}^6 1 \\ & =\frac{6(6+1)(12+1)}{6}-\frac{6(6+1)}{2}+6 \\ & =7 \times 13-3 \times 7+6=91-21+6=76\end{aligned}$
$\begin{aligned} \therefore \sin \left(\sum_{j=1}^6(j+\omega)\left(j+\omega^2\right) \frac{3 \pi}{152}\right) & =\sin \left(\frac{76 \times 3 \pi}{152}\right) \\ & =\sin \frac{3 \pi}{2}=-1\end{aligned}$
$\begin{aligned} \therefore \quad\left(\frac{1-\sqrt{3} i}{2}\right)^{2020}+ & \left(\frac{1+\sqrt{3} i}{2}\right)^{2026} \\ & +\sin \left(\sum_{j=1}^6(j+\omega)\left(j+\omega^2\right) \frac{3 \pi}{512}\right) \\ =-1-1 & =-2\end{aligned}$
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