If omega is an imaginary cube root of unity, | vedclass

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(A) Let the general term be $T_k = (k-1)(k-\omega)(k-\omega^2)$ for $k=2$ to $n$.Since $\omega$ is an imaginary cube root of unity, we have $\omega^3

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the general term be $T_k = (k-1)(k-\omega)(k-\omega^2)$ for $k=2$ to $n$.Since $\omega$ is an imaginary cube root of unity, we have $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$, which implies $\omega + \omega^2 = -1$.Expanding the term: $T_k = (k-1)(k^2 - k(\omega + \omega^2) + \omega^3) = (k-1)(k^2 + k + 1) = k^3 - 1$.The sum is $S = \sum_{k=2}^{n} (k^3 - 1)$.This can be written as $\sum_{k=1}^{n} (k^3 - 1) - (1^3 - 1) = \sum_{k=1}^{n} k^3 - \sum_{k=1}^{n} 1$.Using the standard summation formulas $\sum_{k=1}^{n} k^3 = \frac{n^2(n+1)^2}{4}$ and $\sum_{k=1}^{n} 1 = n$, we get:$S = \frac{n^2(n+1)^2}{4} - n$.

If $\omega$ is an imaginary cube root of unity, then the value of $(2-\omega)(2-\omega^{2}) + 2(3-\omega)(3-\omega^{2}) + \ldots + (n-1)(n-\omega)(n-\omega^{2})$ is

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