If (5+2√6)n=m+f, where n and m are positive integers and 0 ≤ f < 1, then 11−f−f is equal to:
m
We have 5−2√6=1(5+2√6)
Therefore, 0 < 5 - 2√6<1. Let F=(5−2√6)n. Then 0 < F < 1. Also,
m+f+F=(5+2√6)n+(5−2√6)n=2[C05n+C25n−2(2√6)2+C45n−4(2√6)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,
11−f−f=1F−(1−F)=(5+2√6)n−{1−(5−2√6)n}=(5+2√6)n+(5−2√6)n−1
= m + f + F - 1 = m