If `=(2+sqrt(3))^n=I+f,w h e r eIa n dn` positive integers and `0

Explanation

`(2+sqrt(3))^(n) = I + f`
or `I + f = 2^(n) + .^(n)C_(1)2^(n-1) sqrt(3) + .^(n)C_(2)2^(n-2)(sqrt(3))^(2) + .^(n)C_(3)2^(n-3) (sqrt(3))^(3) + "….." (1)`
Now, `0 lt 2 - sqrt(3) - 1`
`rArr 0 (2-sqrt(3))^(n) lt 1`
Let `(2-sqrt(3))^(n) lt 1`
`:. f' = 2^(n) - .^(n)C_(1) 2^(n-1)sqrt(3)+.^(n)C_(2)2^(n-2)(sqrt(3))^(2) - .^(n)C_(3)2^(n-3)(sqrt(3))^(3) + "....." (2)`
Adding (1) and (2), we get
`I + f + f' = 2[2^(n) + .^(n)C_(2)2^(n-2)xxsqrt(3)+"....."]`
or `I + f + f' = "even integer" " "(3)`
Now, `0 lt f lt 1` and `0 lt f' lt 1`
`:. 0 lt f + f' lt 2`
Hence, from (3) we conclude that `f + f'` is an integer between 0 and 2. Therefore,
`f + f' = 1` or `f' = 1 - f " " (4)`
From (3) and (4), we get `I + 1` is an even interger. Therefore, I is an odd integer. Now,
`I + f = (2+sqrt(3))^(n), f' = 1 - f = (2-sqrt(3))^(n)`
`:. (I+f)(1-f) = [(2+sqrt(3))(2-sqrt(3))]^(n) = (4-3)^(n) = I`
`:. (I + f) (1-f) = 1`.