If n is a positive integer and `(5sqrt(5)+11)^(2n+1)=I+f`
where I is an integer and `0 lt f lt 1`, then

Options

A. I is an even integer
B. `(I+f)^(2)` is divisible by `2^(2n+1)`
C. I is divisible by 22.
D. None

Answer

To solve the problem, we need to analyze the expression \((5\sqrt{5} + 11)^{2n + 1}\) and express it in the form \(I + f\), where \(I\) is an integer and \(0 < f < 1\).

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with the expression \((5\sqrt{5} + 11)^{2n + 1}\). Since \(2n + 1\) is an odd integer, we can also consider the conjugate expression \((5\sqrt{5} - 11)^{2n + 1}\).

2. **Using the Binomial Theorem**:
By applying the Binomial Theorem, we can expand both \((5\sqrt{5} + 11)^{2n + 1}\) and \((5\sqrt{5} - 11)^{2n + 1}\):
\[ (5\sqrt{5} + 11)^{2n + 1} = \sum_{k=0}^{2n+1} \binom{2n+1}{k} (5\sqrt{5})^{2n+1-k} (11)^k \]
\[ (5\sqrt{5} - 11)^{2n + 1} = \sum_{k=0}^{2n+1} \binom{2n+1}{k} (5\sqrt{5})^{2n+1-k} (-11)^k \]

3. **Subtracting the Two Expansions**:
When we subtract the second expansion from the first, all terms where \(k\) is odd will cancel out, and we will be left with:
\[ (5\sqrt{5} + 11)^{2n + 1} - (5\sqrt{5} - 11)^{2n + 1} = 2 \sum_{k \text{ even}} \binom{2n+1}{k} (5\sqrt{5})^{2n+1-k} (11)^k \]
This gives us an integer value for \(I\).

4. **Finding \(f\)**:
The term \((5\sqrt{5} - 11)^{2n + 1}\) will yield a very small positive value because \(5\sqrt{5} - 11 < 0\). Therefore, we can denote:
\[ f = (5\sqrt{5} - 11)^{2n + 1} \]
Since \(0 < 5\sqrt{5} - 11 < 1\), it follows that \(0 < f < 1\).

5. **Conclusion**:
We have expressed \((5\sqrt{5} + 11)^{2n + 1}\) in the form \(I + f\) where \(I\) is an integer and \(f\) is a fraction between 0 and 1.

### Final Result:
Thus, we conclude that:
- \(I\) is an even integer.
- \(I\) is divisible by 22.