HMMT February

Algebra Difficulty 5.2 AIME, harder Find the answer

The sequence $a_{1}, a_{2}, a_{3}, \ldots$ of real numbers satisfies the recurrence $a_{n+1}=\frac{a_{n}^{2}-a_{n-1}+2 a_{n}}{a_{n-1}+1}$ . Given that $a_{1}=1$ and $a_{9}=7$ , find $a_{5}$ .

Your answer

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

Solution

Let $b_{n}=a_{n}+1$ . Then the recurrence becomes $b_{n+1}-1=\left(b_{n}^{2}-b_{n-1}\right) / b_{n-1}=b_{n}^{2} / b_{n-1}-1$ , so $b_{n+1}=b_{n}^{2} / b_{n-1}$ . It follows that the sequence $\left(b_{n}\right)$ is a geometric progression, from which $b_{5}^{2}=b_{1} b_{9}=2 \cdot 8=16 \Rightarrow b_{5}= \pm 4$ . However, since all $b_{n}$ are real, they either alternate in sign or all have the same sign (depending on the sign of the progression's common ratio); either way, $b_{5}$ has the same sign as $b_{1}$ , so $b_{5}=4 \Rightarrow a_{5}=3$ .

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