$\text{ Let }\mathrm{R}=(5\sqrt{5}+11{)}^{2\mathrm{n}+1}, \mathrm{f}=\mathrm{R}-[\mathrm{R}]\text{ . Then }\mathrm{Rf}=$

$5\sqrt{5}-11=\frac{(5\sqrt{5}-11)\left(5\sqrt{5}+11\right)}{5\sqrt{5}+11}=\frac{4}{5\sqrt{5}+11}<1\Rightarrow 0<5\sqrt{5}-11<1$

$\begin{array}{l}\text{ Let }\mathrm{h}=(5\sqrt{5}-11{)}^{2\mathrm{n}+1}\text{ . Then }0<\mathrm{h}<1\\ \mathrm{Now} \mathrm{R}-\mathrm{h}=(5\sqrt{5}+11{)}^{2\mathrm{n}+1}-(5\sqrt{5}-11{)}^{2\mathrm{n}+1}\\ =2 \left[{}^{(2\mathrm{n}+1)}\mathrm{C}_1(5\sqrt{5}{)}^{2\mathrm{n}}\left(11\right)+{}^{(2\mathrm{n}+1)}\mathrm{C}_3(5\sqrt{5}{)}^{2\mathrm{n}-2}{\left(11\right)}^3+\dots \right]\end{array}$

                   $$\begin{gathered} = An even positive integer \\ = 2m \left(say\right), m \in N \end{gathered}$$

 f = R – [R]$\Rightarrow$0 < f < 1

$\therefore$$0 < f < 1, 0 < h < 1, f – h is an integer \Rightarrow f = h$

$\therefore \mathrm{Rf}=\mathrm{Rh}=(5\sqrt{5}+11{)}^{2\mathrm{n}+1}.(5\sqrt{5}-11{)}^{2\mathrm{n}+1}=(125-121{)}^{2\mathrm{n}+1}=4^{2\mathrm{n}+1}$