Problem:
The first two terms of a sequence are a1β=1 and a2β=3β1β. For nβ₯1,
an+2β=1βanβan+1βanβ+an+1ββ.
What is β£a2009ββ£?
Answer Choices:
A. 0
B. 2β3β
C. 3β1β
D. 1
E. 2+3β Solution:
Recognize the similarity between the recursion formula given and the trigonometric identity
tan(a+b)=1βtanatanbtana+tanbβ
Also note that the first two terms of the sequence are tangents of familiar angles, namely 4Οβ and 6Οβ. Let c1β=3,c2β=2, and cn+2β=(cnβ+cn+1β)mod12. We claim that the sequence {anβ} satisfies anβ=tan(12Οcnββ). Note that
a1β=1=tan(4Οβ)=tan(12Οc1ββ) and a2β=3β1β=tan(6Οβ)=tan(12Οc2ββ)β
By induction on n, the formula for the tangent of the sum of two angles, and the fact that the period of tanx is Ο,
So the sequence cnβ is periodic with period 24 . Because 2009=24β 83+17, it follows that c2009β=c17β=0. Thus β£a2009ββ£=β£β£β£β£βtan(12Οc17ββ)β£β£β£β£β=0β.