Question

If ω is an imaginary cube root of 1, then the value of 1(2 - ω) (2 - ω2)+ 2(3 - ω) (3 - ω2)+ … + (n-1)(n - ω) (n - ω2) is

• A. $\frac{n(n+1)}{2}-n$
• B. $\frac{n^2(n+1)^2}{4}-n$
• C. $\frac{n(n+1)}{2}+n$
• D. $\frac{n^2(n+1)^2}{4}+n$

Answer 1

Answer: (B)

$\frac{n^2(n+1)^2}{4}-n$

Answer 2

The correct option is (B) : $\frac{n^2(n+1)^2}{4}-n$.