Given that `N = 2^(n)(2^(n+1) -1)` and `2^(n+1) - 1` is a prime number, which of the following is true, where n is a natural number.

Options

A. sum of divisors of N is 2N
B. sum of reciprocals of divisors of N is 1
C. sum of the reciprocals of the divisors of N is 2
D. sum of divisors of N is 4N

Answer

To solve the question, we need to analyze the expression given and the properties of the numbers involved.

### Step-by-step Solution:

1. **Understanding the Expression**:
We start with the expression given:
\[ N = 2^n (2^{n+1} - 1) \]
where \(2^{n+1} - 1\) is a prime number.

2. **Identifying Divisors**:
- The divisors of \(2^n\) are \(1, 2, 2^2, \ldots, 2^n\).
- The divisors of \(2^{n+1} - 1\) (which is prime) are \(1\) and \(2^{n+1} - 1\).

3. **Finding Divisors of \(N\)**:
The divisors of \(N\) can be formed by taking each divisor of \(2^n\) and multiplying it by each divisor of \(2^{n+1} - 1\). This gives us:
- \(1, 2, 2^2, \ldots, 2^n\) (from \(2^n\))
- \(2^{n+1} - 1, 2(2^{n+1} - 1), 2^2(2^{n+1} - 1), \ldots, 2^n(2^{n+1} - 1)\)

4. **Sum of Divisors**:
The sum of the divisors of \(N\) can be calculated as:
\[ \sigma(N) = \sigma(2^n) \cdot \sigma(2^{n+1} - 1) \]
where \(\sigma(2^n) = 1 + 2 + 2^2 + \ldots + 2^n = 2^{n+1} - 1\) (using the formula for the sum of a geometric series).

5. **Calculating \(\sigma(2^{n+1} - 1)\)**:
Since \(2^{n+1} - 1\) is prime, \(\sigma(2^{n+1} - 1) = 1 + (2^{n+1} - 1) = 2^{n+1}\).

6. **Final Calculation**:
Now, substituting back into the sum of divisors:
\[ \sigma(N) = (2^{n+1} - 1) \cdot 2^{n+1} = 2^{n+1}(2^{n+1} - 1) \]

7. **Conclusion**:
The options given in the question are:
- A) Sum of divisors of \(N\) is \(2n\)
- B) Sum of divisors of \(N\) is \(4n\)
- C) Sum of reciprocals of divisors of \(N\) is \(1\)
- D) Sum of reciprocals of divisors of \(N\) is \(2\)

After evaluating the sum of divisors, we find that the correct answer is:
\[ \text{Sum of divisors of } N \text{ is } 2n \text{ (Option A is correct)} \]