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Question:

The value of the expression 1⋅(2−ω)(2−ω²) + 2⋅(3−ω)(3−ω²) +.....+(n−1)(n−ω)(n−ω²), where ω is an imaginary cube root of unity is

n(n+1)²/2 + n

n(n+1)²/2 − n

n(n+1)²/2

None of the above

Solution:

rth term of the given series can be written as:
Tr = (r−1)(r−ω)(r−ω²)
∴ expression = Σr=2ⁿ (r−1)(r−ω)(r−ω²)
= Σr=2ⁿ (r−1)(r² + r + 1)
= Σr=2ⁿ (r³ − 1)
= (2³ + 3³ + … + n³) − (n−1)
= (n(n+1)/2)² − n