Problem:
The first two terms of a sequence are a1=1 and a2=31. For n≥1,
an+2=1−anan+1an+an+1.
What is ∣a2009∣?
Answer Choices:
A. 0
B. 2−3
C. 31
D. 1
E. 2+3 Solution:
Recognize the similarity between the recursion formula given and the trigonometric identity
tan(a+b)=1−tanatanbtana+tanb
Also note that the first two terms of the sequence are tangents of familiar angles, namely 4π and 6π. Let c1=3,c2=2, and cn+2=(cn+cn+1)mod12. We claim that the sequence {an} satisfies an=tan(12πcn). Note that
a1=1=tan(4π)=tan(12πc1) and a2=31=tan(6π)=tan(12πc2)
By induction on n, the formula for the tangent of the sum of two angles, and the fact that the period of tanx is π,