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

The Correct Option is B

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