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HMMT February

Algebra Difficulty 5.1 AIME, harder Find the answer

An infinite sequence of positive real numbers is defined by a0=1 a_{0}=1 and an+2=6an−an+1 a_{n+2}=6 a_{n}-a_{n+1} for n=0,1,2,⋯ n=0,1,2, \cdots Find the possible value(s) of a2007 a_{2007} .

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

The characteristic equation of the linear homogeneous equation is m2+m−6=(m+3)(m−2)=0 m^{2}+m-6=(m+3)(m-2)=0 with solutions m=−3 m=-3 and m=2 m=2 . Hence the general solution is given by an=A(2)n+B(−3)n a_{n}=A(2)^{n}+B(-3)^{n} where A A and B B are constants to be determined. Then we have an>0 a_{n}>0 for n≥0 n \geq 0 , so necessarily B=0 B=0 , and a0=1⇒A=1 a_{0}=1 \Rightarrow A=1 . Therefore, the unique solution to the recurrence is an=2n a_{n}=2^{n} for all n n .

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