If \(\omega\) represents a complex cube root of unity, then…
If \(\omega\) represents a complex cube root of unity, then \(\left(1+\frac{1}{\omega}\right)\left(1+\frac{1}{\omega^2}\right)+\left(2+\frac{1}{\omega}\right)\left(2+\frac{1}{\omega^2}\right)\) \(+\ldots+\left(n+\frac{1}{\omega}\right)\left(n+\frac{1}{\omega^2}\right)=\)
\(\frac{n\left(n^2+1\right)}{3}\)
\(\frac{n\left(n^2+2\right)}{3}\)
\(\frac{n\left(n^2-2\right)}{3}\)
\(\frac{n^2(n-1)}{6}\)
Solution
As \(\omega\) is complex cube root of unity, then
\(\begin{aligned}
& 1+\omega+\omega^2=0 \text { and } \omega^3=1 \quad \ldots (i) \\
& \because\left(r+\frac{1}{\omega}\right)\left(r+\frac{1}{\omega^2}\right)=r^2+r\left(\frac{1}{\omega}+\frac{1}{\omega^2}\right)+\frac{1}{\omega^3} \\
& =r^2+\left(\frac{\omega^2+\omega}{\omega^3}\right) r+\frac{1}{\omega^3}=r^2-r+1 \quad \text { [from Eq. (i)] } \\
& \text { Now, } \sum_{r=1}^n\left(r+\frac{1}{\omega}\right)\left(r+\frac{1}{\omega^2}\right) \\
& =\sum_{r=1}^n\left(r^2-r+1\right)=\frac{n(n+1)(2 n+1)}{6}-\frac{n(n+1)}{2}+n \\
& =\frac{n(n+1)}{6}[2 n+1-3]+n=\frac{n(n+1)}{6}(2 n-2)+n \\
& =\frac{n}{3}\left(n^2-1\right)+n=\frac{n}{3}\left[n^2-1+3\right]=\frac{n\left(n^2+2\right)}{3}
\end{aligned}\)
Hence, option (2) is correct.