The value of the expression `2(1+(1)/(omega))(1+(1)/(omega^(2)))+3(2+(1)/(omega))(2+(1)/(omega^(2)))+4(3+(1)/(omega^()))(3+(1)/(omega^(2)))+.......+(n+1)(n+(1)/(omega^()))(n+(1)/(omega^(2)))` where `omega` is an imaginary cube roots of unity, is:

Options

A. `(n(n^(2)+2))/(3)`
B. `(n(n^(2)-2))/(3)`
C. `(n^(2)(n+1)^(2)+4n)/(4)`
D. None of these

Answer

To solve the expression

\[ S = 2\left(1 + \frac{1}{\omega}\right)\left(1 + \frac{1}{\omega^2}\right) + 3\left(2 + \frac{1}{\omega}\right)\left(2 + \frac{1}{\omega^2}\right) + 4\left(3 + \frac{1}{\omega}\right)\left(3 + \frac{1}{\omega^2}\right) + \ldots + (n+1)\left(n + \frac{1}{\omega}\right)\left(n + \frac{1}{\omega^2}\right) \]

where \(\omega\) is an imaginary cube root of unity, we start by recalling some properties of \(\omega\):

1. \(\omega^3 = 1\)
2. \(1 + \omega + \omega^2 = 0\)

### Step 1: General Term Analysis

The general term of the series can be expressed as:

\[ T_r = (r+1)\left(r + \frac{1}{\omega}\right)\left(r + \frac{1}{\omega^2}\right) \]

### Step 2: Simplifying the General Term

We can simplify \(T_r\):

\[ T_r = (r + 1)\left(r + \frac{1}{\omega}\right)\left(r + \frac{1}{\omega^2}\right) \]

Using the fact that \(\frac{1}{\omega} + \frac{1}{\omega^2} = \frac{\omega + \omega^2}{\omega^3} = \frac{-1}{1} = -1\) and \(\frac{1}{\omega} \cdot \frac{1}{\omega^2} = \frac{1}{\omega^3} = 1\):

\[ T_r = (r + 1)\left(r^2 + \left(\frac{1}{\omega} + \frac{1}{\omega^2}\right)r + \frac{1}{\omega} \cdot \frac{1}{\omega^2}\right) \]

This simplifies to:

\[ T_r = (r + 1)\left(r^2 - r + 1\right) \]

### Step 3: Expanding the General Term

Now, we expand \(T_r\):

\[ T_r = (r + 1)(r^2 - r + 1) = r^3 - r^2 + r + r^2 - r + 1 = r^3 + 1 \]

### Step 4: Summing the Series

Now, we need to sum \(T_r\) from \(r = 1\) to \(n\):

\[ S = \sum_{r=1}^{n} (r^3 + 1) = \sum_{r=1}^{n} r^3 + \sum_{r=1}^{n} 1 \]

The sum of the first \(n\) cubes is given by:

\[ \sum_{r=1}^{n} r^3 = \left(\frac{n(n+1)}{2}\right)^2 \]

And the sum of \(1\) from \(1\) to \(n\) is simply \(n\):

\[ \sum_{r=1}^{n} 1 = n \]

Thus, we have:

\[ S = \left(\frac{n(n+1)}{2}\right)^2 + n \]

### Step 5: Final Expression

Combining the two parts, we get:

\[ S = \frac{n^2(n+1)^2}{4} + n \]

This can be simplified further if needed, but this is the final expression for the value of the given series.