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
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
Updated On: Sep 15, 2024
- \(\frac{n(n+1)}{2}-n\)
- \(\frac{n^2(n+1)^2}{4}-n\)
- \(\frac{n(n+1)}{2}+n\)
- \(\frac{n^2(n+1)^2}{4}+n\)
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The Correct Option is B
Solution and Explanation
The correct option is (B) : \(\frac{n^2(n+1)^2}{4}-n\).
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