For x ∈ R , x ≠ 0 , x ≠ 1 , let f 0 x = 1 1 - x and f n + 1 x = f 0 f n x , n = 0 , 1 , 2 , … . . Then the value of f 100 3 + f 1 2 3 + f 2 3 2 is equal to :

Step-by-step solution

f 1 x = f 0 + 1 x = f 0 f 0 x = 1 1 - 1 1 - x = x - 1 x f 2 x = f 1 + 1 x = f 0 f 1 x = 1 1 - x - 1 x = x f 3 x = f 2 + 1 x = f 0 f 2 x = f 0 x = 1 1 - x f 4 x = f 3 + 1 x = f 0 f 3 x = x - 1 x ∴ f 0 = f 3 = f 6 = … = 1 1 - x f 1 = f 4 = f 7 = … = x - 1 x f 2 = f 5 = f 8 = … = x f 100 3 = 3 - 1 3 = 2 3  , f 1 2 3 = 2 3 - 1 2 3 =   - 1 2 f 2 3 2 = 3 2 ∴ f 100 3 + f 1 2 3 + f 2 3 2 = 5 3