If $ω$ is a complex cube root of unity, then the value of the expression $1 (2 - ω) (2 - ω^{2}) + 2 (3 - ω) (3 - ω^{2}) + . . . + (n - 1) (n - ω) (n - ω^{2})$ is

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

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

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

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

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Solution

The correct option is A $\frac{n^{2} (n + 1)^{2}}{4} - n$
We know,
$(x^{3} - 1) = (x - 1) (x - w) (x - w^{2}) - - - - - - - - - - - - - - - - - - - - - - - - - - - - (1)$
So in the given expression for first term
$1 (2 - w) (2 - w^{2}) = (2 - 1) (2 - w^{2}) (2 - w^{2})$
$= (2^{3} - 1)$ [from equation (1)]
Similarly doing for all terms the expression reduces to
$= 2^{3} - 1 + 3^{3} - 1 + 4^{3} - 1 - - - - - n^{3} - 1$
$= (1^{3} + 2^{3} . . . . n^{3}) - (n - 1) - 1$ [add and subtract 1]
$= \frac{n^{2} (n + 1)^{2}}{4} - n$

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1 \cdot (2 - ω) (2 - ω^{2}) + 2 (3 - ω) (3 - ω^{2}) + . . . . + (n - 1) (n - ω) (n - ω^{2})

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(2 - ω) (2 - ω^{2}) + 2 (3 - ω) (3 - ω^{2}) + . . . + (n - 1) (n - ω) (n - ω^{2})

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