Let R = ( 5 5 + 11 ) 2 n + 1 and f = R - R , where x denotes the greatest integer less than or equal to x ,…

Given, $R={\left(5\sqrt{5}+11\right)}^{2n+1}$ and $f=R-\left[R\right]=\left\{R\right\}$

If $I$ is the integral part of $R$, then $R=I+f={\left(5\sqrt{5}+11\right)}^{2n+1} \dots \left(\mathrm{i}\right)$ and $0<f<1$

Now since $5\sqrt{5}-11=0.18<1$

Let us assume $f_1={\left(5\sqrt{5}-11\right)}^{2n+1} \dots \left(\mathrm{ii}\right)$ and $0<f_1<1$

equation $\left(\mathrm{i}\right)-$equation $\left(\mathrm{ii}\right)$, we get

$I+f-f_1={\left(5\sqrt{5}+11\right)}^{2n+1}-{\left(5\sqrt{5}-11\right)}^{2n+1}$

$\Rightarrow I+f-f_1=2\left[{}^{2n+1}C_1{\left(5\sqrt{5}\right)}^{2n}\left(11\right)+{}^{2n+1}C_3{\left(5\sqrt{5}\right)}^{2n-2}{\left(11\right)}^3+........\right]$

$\Rightarrow I+f-f_1=$ Even integer

$\Rightarrow f-f_1$ must also be an integer

$\Rightarrow f-f_1=0 \left\{\because 0<f<1, 0<f_1<1\right\}$

$\Rightarrow f=f_1$

So, $Rf=R{f}_1={\left(5\sqrt{5}+11\right)}^{2n+1}{\left(5\sqrt{5}-11\right)}^{2n+1}$

$\Rightarrow Rf={\left[\left(5\sqrt{5}+11\right)·\left(5\sqrt{5}-11\right)\right]}^{2n+1}$

$\Rightarrow Rf=4^{2n+1}$