Binomial has a fraction

$(5+2\sqrt{6})^{n}=$ $n \choose 0$ $5^{n}$ + $n \choose 1$ $5^{n-1} \times 2\sqrt6^{1}$ +......... =m+f (fractional part)
$(5-2\sqrt{6})^{n}=$ $n \choose 0$ $5^{n}$ - $n \choose 1$ $5^{n-1} \times 2\sqrt6^{1}$ +......... =p+g (fractional part)
Adding, $(5+2\sqrt{6})^{n}$ + $(5-2\sqrt{6})^{n}$ = all integer terms (irrational terms cancelled). Note that $(5-2\sqrt{6})<1$ . Thus all powers are less than 1 or $p=0$ . ..................................................
Therefore m+f+g = integer. m is also an integer, thus f+g is integer. $0<f+g<2$ implies $f+g=1$
or $g=1-f$ . Also notice that $(5+2\sqrt{6})^{n} \times (5-2\sqrt{6})^{n}=1^{n}=1$ .......................................................
...................................................................................................................................................... $(1-f) \times (m+f)=1 => \frac{1}{1-f}=m+f$ or ................................................................................................ $\frac{1}{1-f}-f =m+f-f=m$ . Hence proved for general $n$

Note that $(5+2\sqrt{6})^n + (5-2\sqrt{6})^n$ is always an integer. You can see this by expanding, or by noting that it lies in ${\mathbb Q}(\sqrt{6}$ ) and is invariant under the automorphism $\sigma(a+b\sqrt{6}) = a - b\sqrt{6}.$

Since $(5-2\sqrt{6})^n$ is a positive real number less than $1,$ we get $\begin{aligned} m &= (5+2\sqrt{6})^n + (5-2\sqrt{6})^n -1 \\ f &= 1-(5-2\sqrt{6})^n \end{aligned}$ Then $\frac1{1-f} - f = \frac1{(5-2\sqrt{6})^n} - (1-(5-2\sqrt{6})^n) = (5+2\sqrt{6})^n + (5-2\sqrt{6})^n - 1 = \fbox{m}.$

Let n=1 (Given n is a Positive integer)

Given $5+2\sqrt { 6 } =m+f\quad (0<f<1)$ and m is a positive integer.

and $5+2\sqrt { 6 } =9.898979...\\ =>\quad m=9\quad and\quad f=0.898979...$

Consider $5-2\sqrt { 6 } =g\quad (0<g<1)$

Now adding both equations.

$10=m+f+g$ and $0<f+g<2$

But $f+g$ is integer. Therefore $f+g=1$ And $m=9$

$g=1-f=5-2\sqrt { 6 } \\ \frac { 1 }{ 1-f } =5+2\sqrt { 6 } \\ and\quad f=2\sqrt { 6 } -4\\ Now\quad \\ \frac { 1 }{ 1-f } -f=5+2\sqrt { 6 } -(2\sqrt { 6 } -4)=9\\ \frac { 1 }{ 1-f } -f=m$