If $\omega$ is a complex cube root of unity, then for any $n1$, $\sum_{r=1}^{n-1} r(r+1-\omega)\left(r+1-\omega^2\right)=$

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

Aicharya · Accepted Answer

We have, $\begin{aligned} & =\sum_{r=1}^{n-1} r(r+1-w)\left(r+1-w^2ight) \ & =\sum_{r=1}^{n-1} r\left[(r+1)^2-(r+1) w^2-w(r+1)+w^3ight] \ & =\sum_{r=1}^{n-1} r\left[(r+1)^2-(r+1)\left(\left(w+w^2ight)+w^3ight)ight] \ & =\sum_{r=1}^{n-1} r\left[(r+1)^2-(r+1)((-1)+1)ight] \ & =\sum_{r=1}^{n-1} r\left(r^2+3 r+3ight) \quad\left[\because w^3=1 	ext { and } 1+w+w^2=0ight] \ & =\sum_{r=1}^{n-1} r^3+3 \sum_{r=1}^{n-1} r^2+3 \sum_{r=1}^{n-1} r \ & =\left(\frac{(n-1) n}{2}ight)^2+\frac{3(n-1)(n)(2 n-1)}{6}+\frac{3(n-1) n}{2} \ & \frac{n^2(n-1)^2}{4}+\frac{n(n-1)(2 n-1)}{2}+\frac{3}{2} n(n-1) \ & =\frac{n(n-1)}{4}[n(n-1)+2(2 n-1)+6] \ & =\frac{n(n-1)}{4}\left[n^2-n+4 n-2+6ight]\end{aligned}$ $=\frac{n(n-1)}{4}\left[n^2+3 n+4ight]$

If $\omega$ is a complex cube root of unity, then for any $n>1$, $\sum_{r=1}^{n-1}…

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