Implicit recursion!
Let the function for all real such that :
Evaluate
The answer is 5.
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1 solution
Note g ( x ) = g ( x + 1 ) + g ( x − 1 ) and g ( x + 1 ) = g ( x ) + g ( x + 2 ) , so g ( x − 1 ) = − g ( x + 2 ) . Therefore, whenever x ≡ y ( m o d 3 ) , ∣ g ( x ) ∣ = ∣ g ( y ) ∣ .
We have g ( 1 7 ) = − g ( 1 7 + 3 ) = − g ( 2 0 ) = − 1 5 , so g ( 1 6 ) = g ( 1 5 ) + g ( 1 7 ) = 2 0 − 1 5 = 5 . Then since 2 0 1 5 2 0 1 5 ≡ 1 ≡ 1 6 ( m o d 3 ) , ∣ g ( 2 0 1 5 2 0 1 5 ) ∣ = ∣ g ( 1 6 ) ∣ = ∣ 5 ∣ = 5 .
(In fact, 2 0 1 5 2 0 1 5 − 1 6 is an odd multiple of 3 , so g ( 2 0 1 5 2 0 1 5 ) = − g ( 1 6 ) = − 5 .)