Question

If ${(5+2\sqrt{6})}^n=m+f$, where $n$ and $m$ are positive integers and $0\le f<1$, then
$\frac{1}{1-f}-f$ is equal to

• A. $\frac{1}{m}$
• B. $m$
• C. $m+\frac{1}{m}$
• D. $m-\frac{1}{m}$

Answer 1

The correct option is B $m$

We have $5+2\sqrt{6}=1/(5-2\sqrt{6})$
Therefore, $0<(5-2\sqrt{6})<1$. Let $F={(5-2\sqrt{6})}^n$. Then $0<F<1$.
Also, $m+f+F={(5+2\sqrt{6})}^n+{(5-2\sqrt{6})}^n$
$=2[C_0{5}^n+C_2{5}^{n-2}{\left(2\sqrt{6}\right)}^2+C_4{5}^{n-4}{\left(2\sqrt{6}\right)}^4+....]=2k$
where $k$ is some positive integer. hence, $f+F=2k-m$ is a positive integer.
Also, $0<f+F<2$. Therefore $f+F=1$.
Now $\frac{1}{1-f}-f=\frac{1}{F}-(1-F)$
$={\{(5+2\sqrt{6})}^n-\{1-{(5-2\sqrt{6})}^n\}$
$={(5+2\sqrt{6})}^n+{(5-2\sqrt{6})}^n-1=m+f+F-1=m$