2016/030) Let $R= (5\sqrt{5} + 11)^{2n+1}$ and $f= R- \lfloor R \rfloor $ where $\lfloor R \rfloor$ is greatest integer functon. Prove that $Rf = 4^{2n+1}$

we have $(5\sqrt{5} - 11) = \frac{(5\sqrt{5} - 11)(5\sqrt{5} - 11)}{5\sqrt{5} + 11}$
$= \frac{125-121}{5\sqrt{5} + 11} = = \frac{4}{5\sqrt{5} + 11} < 1$
$R-r =   (5\sqrt{5} + 11)^{2n+1} -  (5\sqrt{5} - 11)^{2n+1}$
$= \sum \limits_{i=0}^{2n+1}{2n+1\choose i} (5\sqrt{5})^i 11^{2n+1-i} -\sum \limits_{i=0}^{2n+1}{2n+1\choose i} (5\sqrt{5})^i  (-11)^{2n+1-i} $
$= 2\sum \limits_{i=0}^n{2n+1\choose i} (5\sqrt{5})^{2i} 11^{2n+1-1}$
$= 2\sum \limits_{i=0}^n{2n+1\choose i} 125^i 11^{2n+1-1}$ an integer
so $r = R -\lfloor R \rfloor$
So $r = f = (5\sqrt{5} - 11)^{2n+1}$
Hence
$Rf = (5\sqrt{5} +  11)^{2n+1}(5\sqrt{5} - 11)^{2n+1}$
$= ((5\sqrt{5} + 11)(5\sqrt{5} - 11))^{2n+1} = (125-121)^{2n+1} = 4^{2n+1}$