If $\omega$ is a complex cube root of unity, then $\sum_{r=1}^9 r(r+1-\omega)\left(r+1-\omega^2\right)=$

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Aicharya · Accepted Answer

As, $\omega$ is complex cube root of unity, then $\begin{aligned} 1+\omega+\omega^2 & =0 \ \text{and } \omega^3 & =1 \quad\ldots (i) \end{aligned}$ $\begin{aligned} & \because r(r+1-\omega)\left(r+1-\omega^2\right) \ & =r\left[(r+1)^2-\left(\omega+\omega^2\right)(r+1)+\omega^3\right] \ & =r\left[(r+1)^2+(r+1)+1\right] \quad \text { [from Eq. (i)] } \ & =r\left(r^2+3 r+3\right)=r^3+3 r^2+3 r \ & \therefore \sum_{r=1}^9 r(r+1-\omega)\left(r+1-\omega^2\right)=\sum_{r=1}^9\left(r^3+3 r^2+3 r\right) \ & =\left(\frac{9(10)}{2}\right)^2+3 \frac{9(10)(19)}{6}+3 \frac{9 \times 10}{2} \ & \left.=(45)^2+(45 \times 19)+(3 \times 45)=4545+19+3\right] \ & =45 \times 67=3015 \end{aligned}$ Hence, option (4) is correct.

If \(\omega\) is a complex cube root of unity, then \(\sum_{r=1}^9 r(r+1-\omega)\left(r+1-\omega^2\right)=\)

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