If n is an odd integer greater than or equal to 1, then the value of `n^3 - (n-1)^3 + (n-1)^3 - (n-1)^3 + .... + (-1)^(n-1) 1^3`

Options

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

Explanation

To solve the given problem, we need to evaluate the expression:

\[ S = n^3 - (n-1)^3 + (n-1)^3 - (n-2)^3 + (n-2)^3 - (n-3)^3 + \ldots + (-1)^{n-1} 1^3 \]

where \( n \) is an odd integer greater than or equal to 1.

### Step-by-Step Solution:

1. **Identify the Pattern**:
The expression alternates between positive and negative cubes. The first term is \( n^3 \), followed by \( -(n-1)^3 \), then \( (n-1)^3 \), and so on.

2. **Group the Terms**:
We can group the terms in pairs. Notice that each pair consists of a positive cube and a negative cube:
\[ S = n^3 - (n-1)^3 + (n-1)^3 - (n-2)^3 + (n-2)^3 - (n-3)^3 + \ldots + (-1)^{n-1} 1^3 \]
This can be rewritten as:
\[ S = n^3 + \left( - (n-1)^3 + (n-1)^3 \right) + \left( - (n-2)^3 + (n-2)^3 \right) + \ldots + (-1)^{n-1} 1^3 \]

3. **Simplify the Expression**:
The terms \( -(n-1)^3 + (n-1)^3 \) cancel out, and so do \( -(n-2)^3 + (n-2)^3 \), and so on. Thus, we only need to consider the first term:
\[ S = n^3 + \text{(remaining terms cancel out)} \]

4. **Count the Remaining Terms**:
Since \( n \) is odd, the number of terms in the series is \( n \). The last term, when \( n \) is odd, will be \( 1^3 \) with a negative sign.

5. **Final Calculation**:
The only term that remains is \( n^3 \) from the first term, and the last term contributes \( -1^3 \):
\[ S = n^3 - 1 \]

6. **Conclusion**:
Thus, the value of the expression is:
\[ S = n^3 - 1 \]

### Final Answer:
The value of the expression is \( n^3 - 1 \).