Let R =`(sqrt2+1)^(2n+1),ninN` and f= R- [R], where [] denote the greatest integer function, Rf is equal to

Options

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

Answer

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

### Step-by-Step Solution:

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

2. **Express \( f \)**:
\[ f = R - [R] \]
Here, \( f \) represents the fractional part of \( R \).

3. **Calculate \( R \) using the Binomial Theorem**:
\[ R = (\sqrt{2} + 1)^{2n + 1} = \sum_{k=0}^{2n+1} \binom{2n+1}{k} (\sqrt{2})^k (1)^{(2n+1-k)} \]

4. **Calculate \( R' = (\sqrt{2} - 1)^{2n + 1} \)**:
\[ R' = (\sqrt{2} - 1)^{2n + 1} \]
Since \( \sqrt{2} - 1 \) is less than 1, \( R' \) will be a small positive number as \( n \) increases.

5. **Express \( R \) in terms of \( R' \)**:
\[ R = R' + [R] \]
This means \( R \) can be expressed as the sum of an integer part and a small positive number.

6. **Find \( f \)**:
\[ f = R - [R] = R' \quad \text{(since \( R' \) is the fractional part)} \]

7. **Determine the bounds of \( R' \)**:
Since \( R' = (\sqrt{2} - 1)^{2n + 1} \), we know:
\[ 0 < R' < 1 \]

8. **Calculate \( Rf \)**:
\[ Rf = R \cdot f = R \cdot R' = (\sqrt{2} + 1)^{2n + 1} \cdot (\sqrt{2} - 1)^{2n + 1} \]

9. **Simplify \( R \cdot R' \)**:
\[ R \cdot R' = ((\sqrt{2} + 1)(\sqrt{2} - 1))^{2n + 1} = (2 - 1)^{2n + 1} = 1^{2n + 1} = 1 \]

10. **Final Result**:
\[ Rf = 1 \]

### Conclusion:
The value of \( Rf \) is \( 1 \).