If n `in` N such that `(7 + 4 sqrt(3))^(n) = I + F` , then IF is

Options

A. 0
B. 1
C. `7^(2n)`
D. `2^(2n)`

Answer

To solve the problem, we need to find the product \( I \cdot F \) where \( I \) is the integer part and \( F \) is the fractional part of \( (7 + 4\sqrt{3})^n \).

### Step-by-Step Solution:

1. **Understand the Expression**:
We start with the expression \( (7 + 4\sqrt{3})^n \). We denote this as \( R \) for convenience:
\[ R = (7 + 4\sqrt{3})^n \]

2. **Find the Conjugate**:
The conjugate of \( 7 + 4\sqrt{3} \) is \( 7 - 4\sqrt{3} \). We denote this as \( S \):
\[ S = (7 - 4\sqrt{3})^n \]

3. **Calculate the Sum**:
Now, we can calculate the sum of \( R \) and \( S \):
\[ R + S = (7 + 4\sqrt{3})^n + (7 - 4\sqrt{3})^n \]
This sum is an integer because it is the sum of two conjugates.

4. **Determine the Value of \( S \)**:
Since \( 7 - 4\sqrt{3} \) is less than 1 (approximately \( 0.072 \)), as \( n \) increases, \( S \) approaches 0. Thus, for large \( n \), \( S \) becomes negligible.

5. **Express the Integer and Fractional Parts**:
The integer part \( I \) can be expressed as:
\[ I = \lfloor R \rfloor \]
and the fractional part \( F \) is:
\[ F = R - I \]
Since \( S \) is very small, we can approximate:
\[ R + S \approx R \quad \text{(for large } n\text{)} \]

6. **Find the Product \( I \cdot F \)**:
We know that:
\[ R = I + F \]
Therefore, we can express \( I \cdot F \) as:
\[ I \cdot F = I \cdot (R - I) = I \cdot R - I^2 \]

7. **Using the Conjugate**:
Since \( R + S \) is an integer, we can express \( I \) as:
\[ I = R + S - F \]
Thus, we can substitute \( S \) into our expression.

8. **Final Calculation**:
We can find \( I \cdot F \) using:
\[ I \cdot F = R \cdot S \]
Now substituting \( R \) and \( S \):
\[ I \cdot F = (7 + 4\sqrt{3})^n \cdot (7 - 4\sqrt{3})^n = ((7 + 4\sqrt{3})(7 - 4\sqrt{3}))^n = (49 - 48)^n = 1^n = 1 \]

### Conclusion:
Thus, the product \( I \cdot F \) is:
\[ \boxed{1} \]