Let omega be an imaginary cube root of unity. | vedclass

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(A) The general term of the series is $T_r = (r + 1)(r\omega + 1)(r\omega^2 + 1)$ for $r = 1$ to $n$.Using the property $\omega^3 = 1$ and $1 + \omega

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$[\frac{n(n + 1)}{2}]^2 + n$

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(A) The general term of the series is $T_r = (r + 1)(r\omega + 1)(r\omega^2 + 1)$ for $r = 1$ to $n$.Using the property $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$,we have $(r\omega + 1)(r\omega^2 + 1) = r^2\omega^3 + r\omega + r\omega^2 + 1 = r^2 + r(\omega + \omega^2) + 1 = r^2 - r + 1$.Thus,$T_r = (r + 1)(r^2 - r + 1) = r^3 + 1$.The sum is $\sum_{r=1}^n (r^3 + 1) = \sum_{r=1}^n r^3 + \sum_{r=1}^n 1$.Using the formula $\sum_{r=1}^n r^3 = [\frac{n(n + 1)}{2}]^2$,the sum is $[\frac{n(n + 1)}{2}]^2 + n$.

Let $\omega$ be an imaginary cube root of unity. Then the value of $2(\omega + 1)(\omega^2 + 1) + 3(2\omega + 1)(2\omega^2 + 1) + \dots + (n + 1)(n\omega + 1)(n\omega^2 + 1)$ is

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