Let $R={(5\sqrt{5}+11)}^{2n+1}$ and $f=R-\left[R\right]$ where $\left[ \right]$ denotes the greatest integral function, then $Rf=$

$\begin{array}{l}R={(5\sqrt{5}+11)}^{2n+1}\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}f=R-\left[R\right]\\ Rf=4^{2n+1}\\ R={(5\sqrt{5}+11)}^{2n+1}=I+f,0<f<1\cdots \left(1\right)\\ F={(5\sqrt{5}-11)}^{2n+1},0<F<1\Rightarrow -1<-F<0\cdots \left(2\right)\\ eq\left(1\right)-eq\left(2\right)\\ I+f-F={(5\sqrt{5}+11)}^{2n+1}-{(5\sqrt{5}-11)}^{2n+1}\\ =2[{}^{2n+1}{\text{C}}_0{\left(5\sqrt{5}\right)}^{2n+1}+{}^{2n+1}{\text{C}}_2{\left(5\sqrt{5}\right)}^{2n-1}.{\left(11\right)}^2+\cdots ]\\ =2\left[\text{Integers}\right]\end{array}$

$I+f-F=$Even integer

I is an integer. $f-F$ is also an integer

$f-F=0\Rightarrow f=F$

$Rf=RF={(5\sqrt{5}+11)}^{2n+1}.{(5\sqrt{5}-11)}^{2n+1}$

$=4^{2n+1}$