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 given 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 can break it down step by step.

### Step 1: Understanding the properties of \(\omega\)

The cube roots of unity satisfy the following properties:
1. \(\omega^3 = 1\)
2. \(1 + \omega + \omega^2 = 0\)

### Step 2: Simplifying the \(r\)-th term

The \(r\)-th term can be expressed as:

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

Calculating \(\left(1 + \frac{1}{\omega}\right)\left(1 + \frac{1}{\omega^2}\right)\):

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

Since \(\omega \cdot \omega^2 = \omega^3 = 1\), we have:

\[ = 1 + \frac{1}{\omega} + \frac{1}{\omega^2} + 1 = 2 + \frac{1}{\omega} + \frac{1}{\omega^2} \]

Now, using the property \(1 + \omega + \omega^2 = 0\), we can express \(\frac{1}{\omega} + \frac{1}{\omega^2}\) as:

\[ \frac{1}{\omega} + \frac{1}{\omega^2} = \frac{\omega + 1}{\omega^2} = -1 \quad (\text{since } \omega + \omega^2 = -1) \]

Thus,

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

### Step 3: Substituting back into the \(r\)-th term

Now we have:

\[ T_r = r(r + 1) \cdot 1 = r(r + 1) \]

### Step 4: Summing the series

The entire expression \(S\) can now be rewritten as:

\[ S = \sum_{r=1}^{n} r(r + 1) = \sum_{r=1}^{n} (r^2 + r) \]

This can be separated into two sums:

\[ S = \sum_{r=1}^{n} r^2 + \sum_{r=1}^{n} r \]

Using the formulas for these sums:

1. \(\sum_{r=1}^{n} r^2 = \frac{n(n + 1)(2n + 1)}{6}\)
2. \(\sum_{r=1}^{n} r = \frac{n(n + 1)}{2}\)

### Step 5: Final expression for \(S\)

Putting it all together:

\[ S = \frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2} \]

To combine these, we can find a common denominator:

\[ S = \frac{n(n + 1)(2n + 1)}{6} + \frac{3n(n + 1)}{6} = \frac{n(n + 1)(2n + 1 + 3)}{6} = \frac{n(n + 1)(2n + 4)}{6} \]

Simplifying further gives:

\[ S = \frac{n(n + 1)(n + 2)}{3} \]

### Final Answer

Thus, the value of the expression is:

\[ \frac{n(n + 1)(n + 2)}{3} \]