The value of the expression 1.(2 - omega )(2 | vedclass

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(B) The $r^{th}$ term of the given series is $T_r = r[(r + 1) - \omega][(r + 1) - \omega^2]$.Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omeg

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$\frac{1}{4}(n - 1)n({n^2} + 3n + 4)$

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(B) The $r^{th}$ term of the given series is $T_r = r[(r + 1) - \omega][(r + 1) - \omega^2]$.Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$ and $\omega^3 = 1$.$T_r = r[(r + 1)^2 - (\omega + \omega^2)(r + 1) + \omega^3] = r[(r + 1)^2 + (r + 1) + 1] = r[r^2 + 2r + 1 + r + 1 + 1] = r(r^2 + 3r + 3) = r^3 + 3r^2 + 3r$.The sum of the series is $S = \sum_{r=1}^{n-1} (r^3 + 3r^2 + 3r)$.$S = \sum_{r=1}^{n-1} r^3 + 3 \sum_{r=1}^{n-1} r^2 + 3 \sum_{r=1}^{n-1} r$.$S = [\frac{(n-1)n}{2}]^2 + 3 \cdot \frac{(n-1)n(2n-2+1)}{6} + 3 \cdot \frac{(n-1)n}{2}$.$S = \frac{(n-1)^2 n^2}{4} + \frac{(n-1)n(2n-1)}{2} + \frac{3(n-1)n}{2}$.$S = \frac{(n-1)n}{4} [(n-1)n + 2(2n-1) + 6] = \frac{(n-1)n}{4} [n^2 - n + 4n - 2 + 6] = \frac{1}{4}(n-1)n(n^2 + 3n + 4)$.

The value of the expression $1.(2 - \omega )(2 - {\omega ^2}) + 2.(3 - \omega )(3 - {\omega ^2}) + ....... + (n - 1).(n - \omega )(n - {\omega ^2}),$ where $\omega$ is an imaginary cube root of unity,is

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