If `(6sqrt(6)+14 )^(2n+1)` = [N]+F and F=N -[N] , where [.] denotes greatest interger function then NF is equal to

Options

A. `20^(2n+1)`
B. an even interger
C. odd interger
D. `40^(2n+1)`

Explanation

Since `(6sqrt(6) +14)^(2n+1)= [N]+F`
Let us same that f `=(6sqrt(6)14)^(2n+1)` : where of ` 0 le f lt 1`
Now ,[N] +F-f`=6(sqrt(6)-14)^(2n+1)-(6 sqrt(6)-14)^(2n+1)`
`=2[""^(2n+1)C_(1)(6sqrt(6))^2n (14)+^(2n+1)C_(3)(6sqrt(6))^(2n-2)(14)^(3)+..]`
`rArr [N] +F =` even interger
Now ` j lt F lt 1 and 0 lt f lt 1`
So -1 lt F- f lt 1 and F -f is an integer so it can only be zero
Thus NF=`(6sqrt(6)+14)^(2n+1)(6sqrt(6)-14)^(2n +1)=20^(2n+1)`