Let `R=(5sqrt(5)+11)^(2n+1)a n df=R-[R]w h e r e[]` denotes the greatest integer function, prove that `Rf=4^(2n+1)`

Explanation

Here `f = R - [R]` is the fractional part of R. Thus, if I is the integral part of R, then
`R = I + f = (5sqrt(5)+11)^(2n+1)` , and `0 lt f lt 1`
Let `f' = (5sqrt(5)-11)^(2n+1)`
Then `0 lt f' lt 1`
Now, `I + f - f' = (5sqrt(5) +11)^(2n+1) - (5sqrt(5) + 11)^(2n+1)`
`= 2[.^(2n+1)C_(1)(5sqrt(5))^(2n) xx 11 + .^(2n+1)C_(3)(5sqrt(5))^(2n-2) xx 11^(3) + "....."]`
= an even integer " "(1)
`rArr f - f'` must also be an integer `rArr f - f' = 0, ( :' 0 lt lt f lt 1, 0 lt f' lt 1)`
`rArr f = f'`
`:. Rf = Rf' = (5sqrt(5)+11)^(2n+1) (5sqrt(5)-11)^(2n+1)`
`= (125-121)^(2n+1) = 4^(2n+1)`