Problem: Let f(x) be a function such that f(1)=1, f(2)=2 and f(x+2)=f(x+1)−f(x). Find f(2016).
Solution
Solution: f(1)=1f(2)=2f(3)=f(2)−f(1)=2−1=1f(4)=f(3)−f(2)=1−2=−1f(5)=f(4)−f(3)=−1−1=−2f(6)=f(5)−f(4)=−2−(−1)=−1f(7)=f(6)−f(5)=−1−(−2)=1f(8)=f(7)−f(6)=1−(−1)=2 Observe that this pattern will repeat itself every six, and thus, f(2016)=f(6)=−1.
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Source: MathNet,
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