If `(4+sqrt(15))^n=I+f,w h r en` is an odd natural number, `I` is an integer and `0

Options

A. I is an odd integer
B. I is an even nteger
C. `(I + f) (I - f) = 1`
D. `I - f = (4 - sqrt(15))^(n)`

Answer

A, C, D

Explanation

`I + f = (4+sqrt(5))^(n)`
Let `f'=(4+sqrt(15))^(n)`, then `0 lt f' lt 1`
`I+f = .^(n)C_(0)4^(n) + .^(n)C_(1)4^(n-1)sqrt(15)+.^(n)C_(2)4^(n-2)15+.^(n)C_(3)4^(n-3)(sqrt(5))^(3)+"......"`
`f' = .^(n)C_(0)4^(n)-.^(4)C_(1)4^(n-1)sqrt(15)+.^(n)C_(2)4^(n-2).15-.^(n)C_(3) 4^(n-3)(sqrt(15))^(3) + "......"`
`:. I + f + f' = 2(.^(n)C_(0)4^(n)+.^(n)C_(2)4^(n-2)xx15+"....")` = even integer
`:'0 lt f+ f' lt 2 rArr f + f' rArr 1 - f = f'`
Thus, I is odd integer, Now,
`I + f = f' = (4-sqrt(15))^(n)`
`(1+f) (1-f) = (I+f)f' = 1`