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

Options

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

Answer

To solve the expression

\[ (1 + \frac{1}{\omega})(1 + \frac{1}{\omega^2}) + (2 + \frac{1}{\omega})(2 + \frac{1}{\omega^2}) + (3 + \frac{1}{\omega})(3 + \frac{1}{\omega^2}) + \ldots + (n + \frac{1}{\omega})(n + \frac{1}{\omega^2}), \]

where \(\omega\) is an imaginary cube root of unity, we can follow these steps:

### Step 1: Understanding \(\omega\)
The cube roots of unity are given by:
\[ \omega = e^{2\pi i / 3}, \quad \omega^2 = e^{-2\pi i / 3}, \quad \text{and } \omega^3 = 1. \]
Also, we know that:
\[ 1 + \omega + \omega^2 = 0. \]

### Step 2: Simplifying the First Term
Let's start with the first term:
\[ (1 + \frac{1}{\omega})(1 + \frac{1}{\omega^2}). \]
Expanding this gives:
\[ 1 + \frac{1}{\omega} + \frac{1}{\omega^2} + \frac{1}{\omega \cdot \omega^2}. \]
Since \(\omega \cdot \omega^2 = \omega^3 = 1\), we have:
\[ 1 + \frac{1}{\omega} + \frac{1}{\omega^2} + 1 = 2 + \frac{1}{\omega} + \frac{1}{\omega^2}. \]
Using \( \frac{1}{\omega} + \frac{1}{\omega^2} = \frac{\omega + 1}{\omega^2} = -1 \) (since \(1 + \omega + \omega^2 = 0\)), we find:
\[ 2 - 1 = 1. \]

### Step 3: Generalizing the \(k\)-th Term
Now, consider the \(k\)-th term:
\[ (k + \frac{1}{\omega})(k + \frac{1}{\omega^2}). \]
Expanding this gives:
\[ k^2 + k(\frac{1}{\omega} + \frac{1}{\omega^2}) + \frac{1}{\omega \cdot \omega^2}. \]
Again, since \(\frac{1}{\omega} + \frac{1}{\omega^2} = -1\) and \(\frac{1}{\omega \cdot \omega^2} = 1\), we have:
\[ k^2 - k + 1. \]

### Step 4: Summing the Series
Now we need to sum this from \(k = 1\) to \(n\):
\[ \sum_{k=1}^{n} (k^2 - k + 1). \]
This can be split into three separate sums:
\[ \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1. \]

### Step 5: Using Known Formulas
Using the formulas for the sums:
- \(\sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6}\),
- \(\sum_{k=1}^{n} k = \frac{n(n + 1)}{2}\),
- \(\sum_{k=1}^{n} 1 = n\).

We can substitute these into our expression:
\[ \frac{n(n + 1)(2n + 1)}{6} - \frac{n(n + 1)}{2} + n. \]

### Step 6: Simplifying the Expression
Now, we can simplify:
1. Find a common denominator (which is 6):
\[ \frac{n(n + 1)(2n + 1)}{6} - \frac{3n(n + 1)}{6} + \frac{6n}{6}. \]
2. Combine the fractions:
\[ \frac{n(n + 1)(2n + 1) - 3n(n + 1) + 6n}{6}. \]

### Step 7: Final Simplification
Expanding and simplifying the numerator:
\[ n(n + 1)(2n + 1) - 3n(n + 1) + 6n = n(n + 1)(2n + 1 - 3) + 6n = n(n + 1)(2n - 2) + 6n. \]
This simplifies further to:
\[ n(n + 1)(2(n - 1)) + 6n = 2n(n + 1)(n - 1) + 6n. \]
Thus, the final result is:
\[ \frac{n(n^2 + 2)}{3}. \]

### Final Answer
The value of the expression is:
\[ \frac{n(n^2 + 2)}{3}. \]