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Mathematics Question on Complex Numbers and Quadratic Equations

If ω\omega is an imaginary cube root of unity, then the value of (2−ω)(2−ω2)+2(3−ω)(3−ω2)+.....+(n−1)(n−ω)(n−ω2)\left(2-\omega\right)\left(2-\omega^{2}\right)+2\left(3-\omega\right)\left(3-\omega^{2}\right)+.....+\left(n-1\right)\left(n-\omega\right)\left(n-\omega^{2}\right)

A

n24(n+1)2−n\frac{n^{2}}{4}\left(n+1\right)^{2}-n

B

n24(n+1)2+n\frac{n^{2}}{4}\left(n+1\right)^{2}+n

C

n24(n+1)2\frac{n^{2}}{4}\left(n+1\right)^{2}

D

n24(n+1)−n\frac{n^{2}}{4}\left(n+1\right)-n

Answer

n24(n+1)2−n\frac{n^{2}}{4}\left(n+1\right)^{2}-n

Explanation

Solution

∑r=2n(r−1)(r−ω2)=∑r=2n(r3−1)=\displaystyle \sum_{r=2}^n\left(r-1\right)\left(r-\omega^{2}\right)=\displaystyle \sum_{r=2}^n(r^3-1)= [n2(n+1)24−1]−(n−1)=n2(n+1)24−n\left[\frac{n^{2}\left(n+1\right)^{2}}{4}-1\right]-\left(n-1\right)=\frac{n^{2}\left(n+1\right)^{2}}{4}-n