If `P=(6sqrt(6)+14)^(2n+1)` and `f=P-[P]`, where [ ] denotes the greatest integer function, then Pf equals

Options

A. `20^(n)`
B. `20^(2n)`
C. `20^(2n-1)`
D. none of these

Answer

To solve the problem, we start with the given expression for \( P \) and \( f \):

1. **Given:**
\[ P = (6\sqrt{6} + 14)^{2n + 1} \]
\[ f = P - [P] \]
Here, \( [P] \) denotes the greatest integer function, which means \( f \) represents the fractional part of \( P \).

2. **Understanding \( f \):**
Since \( f = P - [P] \), we can rewrite it as:
\[ f = P - \text{(the greatest integer less than or equal to } P\text{)} \]
This implies that \( f \) is the fractional part of \( P \).

3. **Finding \( P \cdot f \):**
We need to calculate \( P \cdot f \):
\[ P \cdot f = P \cdot (P - [P]) = P^2 - P \cdot [P] \]

4. **Calculating \( P^2 \):**
We can simplify \( P^2 \):
\[ P^2 = \left( (6\sqrt{6} + 14)^{2n + 1} \right)^2 = (6\sqrt{6} + 14)^{2(2n + 1)} = (6\sqrt{6} + 14)^{4n + 2} \]

5. **Calculating \( P \cdot [P] \):**
To find \( P \cdot [P] \), we need to express \( [P] \). Since \( P \) is a positive number, we can express \( [P] \) as:
\[ [P] = \lfloor (6\sqrt{6} + 14)^{2n + 1} \rfloor \]
However, for the purpose of this calculation, we can focus on the fact that \( P \cdot [P] \) will be very close to \( P^2 \).

6. **Using the Conjugate:**
We can use the conjugate of \( (6\sqrt{6} + 14) \):
\[ (6\sqrt{6} - 14)^{2n + 1} \]
This is because the product of \( (a + b)(a - b) = a^2 - b^2 \).

7. **Final Calculation:**
Therefore, we can express \( P \cdot f \) as:
\[ P \cdot f = (6\sqrt{6} + 14)^{2n + 1} \cdot (6\sqrt{6} - 14)^{2n + 1} \]
\[ = ((6\sqrt{6})^2 - 14^2)^{2n + 1} \]
\[ = (216 - 196)^{2n + 1} = (20)^{2n + 1} \]

8. **Conclusion:**
Thus, the final result is:
\[ P \cdot f = 20^{2n + 1} \]