Let `(6sqrt(6)+14)^(2n+1)=R`, if R be the fractional part of R, then prove that `Rf=20^(2n+1)`

Explanation

`(6sqrt(6) + 14)^(2n + 1) ` can be written as ` (sqrt(216 ) + 14)^(2n+1)` and
given that f = R - [R]
and ` R = (6 sqrt(6) + 14)^(2n +1) = (sqrt(216) + 14)^(2n +1)`
` therefore p[R] + f = 9sqrt(216) + 14_^(2n +1)` …(i)
` 0 le f lt 1 ` ...(ii)
Let ` f ' = (sqrt(216) - 14 )^(2n +1)` ...(iii)
` 0 lt f' lt 1` ...(iv)
On subtracting . Eq (iii) from Eq. (i) , we get
` [R] + f - f' = (sqrt(216 ) + 14)^(2n + 1) - (sqrt(216) - 14)^(2n + 1)`
[R] + 0 = 2p , ` AA p in ` N = Even integer [ from theorem 1]
` therefore f- f' = 0 " or" f = f' `
Now , ` Rf= Rf' = (sqrt(216) + 14 )^(2n + 1) (sqrt(216) - 14) ^(2n +1)`
`= (216 - 196)^(2n +1) = (20)^(2n +1)` .