If omega is a complex cube root of unity, the | vedclass

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(A) The general term of the series is $T_k = (k + 1)(k + \frac{1}{\omega})(k + \frac{1}{\omega^2})$ for $k = 1$ to $n$.Since $\omega^3 = 1$ and $1 + \

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

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

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

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