If R = `(sqrt(2) + 1)^(2n+1) and f = R - [R]`, where [ ]
denote the greatest integer function, then [R] equal (a) `f+1/f` (b) `f-1/f` (c) `1/f-f` (d) None of these

Options

A. `f+1/f`
B. `f-1/f`
C. `1/f-f`
D. None of these

Answer

To solve the problem, we need to find the value of \([R]\) where \(R = (\sqrt{2} + 1)^{2n + 1}\) and \(f = R - [R]\).

### Step-by-Step Solution:

1. **Define \(R\)**:
\[ R = (\sqrt{2} + 1)^{2n + 1} \]

2. **Define \(g\)**:
We can also express \(g\) as:
\[ g = (\sqrt{2} - 1)^{2n + 1} \]

3. **Relate \(R\) and \(g\)**:
Notice that:
\[ R - g = (\sqrt{2} + 1)^{2n + 1} - (\sqrt{2} - 1)^{2n + 1} \]
This expression can be simplified using the binomial theorem.

4. **Using the Binomial Theorem**:
Expanding both terms using the binomial theorem:
\[ R = \sum_{k=0}^{2n+1} \binom{2n+1}{k} (\sqrt{2})^k (1)^{2n+1-k} \]
\[ g = \sum_{k=0}^{2n+1} \binom{2n+1}{k} (\sqrt{2})^k (-1)^{2n+1-k} \]

5. **Combine the two expansions**:
The odd indexed terms in \(R\) and \(g\) will cancel out, leaving us with:
\[ R - g = \text{(even integer)} \]

6. **Determine \(f\)**:
Since \(f = R - [R]\), we can express \(f\) as:
\[ f = R - g + g - [R] \]
Since \(g\) is a very small positive number (as \((\sqrt{2} - 1) < 1\)), we can say:
\[ f \approx R - [R] \]

7. **Find \([R]\)**:
From the relationship \(R - g\) being an even integer, we can conclude that:
\[ [R] = f - \frac{1}{f} \]

8. **Final Expression**:
Thus, we can express \([R]\) as:
\[ [R] = f - \frac{1}{f} \]

### Conclusion:
The value of \([R]\) is:
\[ \boxed{f - \frac{1}{f}} \]