Find the sum `1xx(2-omega)xx(2-omega^(2))+2xx(-3-omega)xx(3-omega^(2))+….+(n-1)xx(n-omega)xx(n-omega^(2))`, where `omega` is an imaginary cube root of unity.

Options

A. `{(n(n+1))/(2)}^(2)`
B. `{(n(n+1))/(2)}^(2)-n`
C. `{(n(n+1))/(2)}^(2)+n`
D. none of these

Answer

B

Explanation

We have,
`(z-1)(z-omega)(z-omega^(2))=z^(3)-1`
`therefore 1(2-omega)(2-omega^(2))+2(3-omega)(3-omega^(2))+….++(n-1)(n-omega)(n-omega^(2))`
`=sum_(r=2)^(n)(r-1)(r-omega)(r-omega^(2))`
`=sum_(r=2)^(n) (r^(3)-1)=sum_(r=2)^(n) r^(3) -sum_(r=2)^(n)1=(sum_(r=2)^(n)r^(3))-1-(sum_(r=2)^(n)1)`
`={(n(n+1))/(2)}^(2)-1-(n-1)={(n(n+1))/(2)}^(2)-n`