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 $5-2\sqrt{6}=\frac{1}{(5+2\sqrt{6})}$
Therefore, 0 < 5 - $2\sqrt{6}<1.LetF=(5-2\sqrt{6}{)}^n$. Then 0 < F < 1. Also,
$m+f+F=(5+2\sqrt{6}{)}^n+(5-2\sqrt{6}{)}^n=2[C_0{5}^n+C_2{5}^{n-2}(2\sqrt{6}{)}^2+C_4{5}^{n-4}(2\sqrt{6}{)}^4+\cdots ]=2k$
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