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(B) Since $\omega$ is a complex cube root of unity,we have $1+\omega+\omega^2=0$ and $\omega^3=1$. Consider the general term $T_r = \left(r+\frac{1}{\

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$\frac{n(n^2+2)}{3}$

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(B) Since $\omega$ is a complex cube root of unity,we have $1+\omega+\omega^2=0$ and $\omega^3=1$. Consider the general term $T_r = \left(r+\frac{1}{\omega}\right)\left(r+\frac{1}{\omega^2}\right)$. Expanding this,we get $T_r = r^2 + r\left(\frac{1}{\omega} + \frac{1}{\omega^2}\right) + \frac{1}{\omega^3}$. Since $\frac{1}{\omega} = \omega^2$ and $\frac{1}{\omega^2} = \omega$,we have $\frac{1}{\omega} + \frac{1}{\omega^2} = \omega^2 + \omega = -1$. Also,$\frac{1}{\omega^3} = 1$. Thus,$T_r = r^2 - r + 1$. The sum is $\sum_{r=1}^n (r^2 - r + 1) = \sum_{r=1}^n r^2 - \sum_{r=1}^n r + \sum_{r=1}^n 1$. Using standard summation formulas: $\frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} + n$. $= \frac{n(n+1)(2n+1) - 3n(n+1) + 6n}{6} = \frac{n[(n+1)(2n+1) - 3(n+1) + 6]}{6}$. $= \frac{n[2n^2 + 3n + 1 - 3n - 3 + 6]}{6} = \frac{n[2n^2 + 4]}{6} = \frac{2n(n^2+2)}{6} = \frac{n(n^2+2)}{3}$.

If $\omega$ represents a complex cube root of unity,then $\left(1+\frac{1}{\omega}\right)\left(1+\frac{1}{\omega^2}\right)+\left(2+\frac{1}{\omega}\right)\left(2+\frac{1}{\omega^2}\right)+\ldots+\left(n+\frac{1}{\omega}\right)\left(n+\frac{1}{\omega^2}\right)=$

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