The value of the expression `2 (1 + omega) (1 + omega^(2)) + 3(2 omega + 1) (2 omega^(2) + 1) + 4(3 omega + 1) (3 omega^(2) + 1)+`
…`+ (n + 1) (n omega + 1) (n omega^(2) + 1)` is

Options

A. `[(n(n + 1))/(2)]^(2)`
B. `[(n(n + 1))/(2)]^(2)+n`
C. `[(n(n + 1))/(2)]^(2)-n`
D. none of these

Answer

To solve the expression \(2 (1 + \omega)(1 + \omega^2) + 3(2\omega + 1)(2\omega^2 + 1) + 4(3\omega + 1)(3\omega^2 + 1) + \ldots + (n + 1)(n\omega + 1)(n\omega^2 + 1)\), we will follow these steps:

### Step 1: Identify the General Term
The general term of the expression can be written as:
\[ T_n = (n + 1)(n\omega + 1)(n\omega^2 + 1) \]

### Step 2: Expand the General Term
Now, we will expand \(T_n\):
\[ T_n = (n + 1)((n\omega)(n\omega^2) + n\omega + n\omega^2 + 1) \]
Since \(\omega^3 = 1\), we know that \(\omega^2 + \omega + 1 = 0\). Therefore, we can simplify:
\[ T_n = (n + 1)(n^2\omega^3 + n\omega + n\omega^2 + 1) = (n + 1)(n^2 + n\omega + n\omega^2 + 1) \]
Using the identity \(1 + \omega + \omega^2 = 0\):
\[ T_n = (n + 1)(n^2 + 1 - n) = (n + 1)(n^2 - n + 1) \]

### Step 3: Simplify Further
Now we can simplify:
\[ T_n = (n + 1)(n^2 - n + 1) = (n + 1)(n^2 - n + 1) \]

### Step 4: Summation of the General Term
To find the total value of the expression, we need to sum \(T_n\) from \(n = 1\) to \(n\):
\[ \text{Total} = \sum_{n=1}^{N} T_n = \sum_{n=1}^{N} (n + 1)(n^2 - n + 1) \]

### Step 5: Break Down the Summation
We can break down the summation:
\[ \sum_{n=1}^{N} (n + 1)(n^2 - n + 1) = \sum_{n=1}^{N} (n^3 + n^2 - n^2 - n + n + 1) = \sum_{n=1}^{N} (n^3 + 1) \]

### Step 6: Use Known Summation Formulas
Using the formula for the sum of cubes:
\[ \sum_{n=1}^{N} n^3 = \left(\frac{N(N + 1)}{2}\right)^2 \]
And the sum of constants:
\[ \sum_{n=1}^{N} 1 = N \]
Thus, we have:
\[ \text{Total} = \left(\frac{N(N + 1)}{2}\right)^2 + N \]

### Final Result
The value of the expression is:
\[ \frac{N(N + 1)}{2}^2 + N \]