Let `(6sqrt(6)+14)^(2n+1)=R`, if f be the fractional part of R, then prove that `Rf=20^(2n+1)`

Explanation

To solve the problem, we need to prove that if \( R = (6\sqrt{6} + 14)^{2n+1} \) and \( f \) is the fractional part of \( R \), then \( Rf = 20^{2n+1} \).

### Step-by-step Solution:

1. **Define \( R \) and \( f \)**:
\[ R = (6\sqrt{6} + 14)^{2n+1} \]
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