A sequence of real numbers is defined recursively as follows:
Given that and , find .
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Direct substitution in the recurrence relation f n = f n − 2 f n − 1 + 1 , n = 0 , 1 , 2 , ⋯ gives the following :-
f 0 f 1 f 2 f 3 f 4 f 5 f 6 f 7 f 8 f 9 f 1 0 f 1 1 f 1 2 f 1 3 f 1 4 f 1 5 ⋯ = 2 4 . 8 = 3 6 . 9 = 1 . 5 2 8 2 3 = 0 . 0 6 8 5 1 5 6 = 0 . 6 9 9 1 8 7 = 2 4 . 8 = 3 6 . 9 = 1 . 5 2 8 2 3 = 0 . 0 6 8 5 1 5 6 = 0 . 6 9 9 1 8 7 = 2 4 . 8 = 3 6 . 9 = 1 . 5 2 8 2 3 = 0 . 0 6 8 5 1 5 6 = 0 . 6 9 9 1 8 7 = 2 4 . 8 = ⋯
The above sequence has cycle with length = 5
f 4 4 4 5 = f 4 4 4 5 % 5 = f 0 = 2 4 . 8
Answer = 1 0 × 2 4 . 8 = 2 4 8