To solve this problem, we need to understand the properties of the imaginary cube roots of unity. The cube roots of unity are \(1, \omega, \omega^2\), where \(\omega\) is a non-real root. They satisfy the equation:
\(\omega^3 = 1\) and \(1 + \omega + \omega^2 = 0\).
Let's analyze the given expression:
\(2(1+\omega)(1+\omega^2) + 3(2\omega+1)(2\omega^2+1) + \cdots + (n+1)(n\omega+1)(n\omega^2+1)\).
First, calculate \((1+\omega)(1+\omega^2)\):
We know:
\((1+\omega)(1+\omega^2) = 1 + \omega + \omega^2 + \omega \cdot \omega^2 = 1 + \omega + \omega^2 + \omega^3 = 1 + 0 + 1 = 1\).
The calculation shows that \((1+\omega)(1+\omega^2) = 1\) since \(\omega^3 = 1\) and \(\omega + \omega^2 = -1\).
Now look at the general form:
\((k\omega+1)(k\omega^2+1) = k^2\omega \cdot \omega^2 + k(\omega + \omega^2) + 1 = k^2 + 1 - k\).
\(= (k-1)^2\)
Now, substitute back into the original expression:
Using \(k = 1, 2, ..., n\), the sum becomes:
\(\sum_{k=1}^{n} (k+1)(k-1)^2\) which simplifies as:
Each term for a specific \(k\) becomes:
\((k+1)(k-1)^2 = (k+1)(k^2 - 2k + 1)\)
So, for each value of \(k\), it's:
\(= (k^3 - 2k^2 + k^2 - 2k + k)\)
\(= k^3 - k\),\) perform the summation:
The complete expression evaluates to:
\(= \left(\sum_{k=1}^{n} k^3\right) - \left(\sum_{k=1}^{n} k\right) = \left(\frac{n(n+1)}{2}\right)^2 - \frac{n(n+1)}{2} + n\).
Thus, the entire expression simplifies to:
\(= \left(\frac{n(n+1)}{2}\right)^2 + n\).
The answer matches with option:
\(\left[\frac{n(n+1)}{2}\right]^2 + n\)