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