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Problem 1117

AIME late

Algebra Difficulty 5.0 Prove it Romanian Mathematical Olympiad · Romania

Let $(a_n)_{n \ge 0}$ be a sequence of positive real numbers such that
$\sum_{k=0}^{n} C_{n}^{k} a_{k} a_{n-k} = a_{n}^{2}, \text{ for any } n \ge 0.$
Prove that $(a_n)_{n \ge 0}$ is a geometrical sequence.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let $a_0 = a$ . It is obvious that $a_1 = 2a$ , and, for $n = 2$ ,
$a_2^2 - 2aa_2 - 8a^2 = 0,$
implying $a_2 = 4a$ , for $a_2 > 0$ .
Use induction on $n$ to prove that $a_n = 2^n a$ . Assume that $a_k = 2^k a$ , for all $k$ , $0 \le k \le n$ , to prove $a_{n+1} = 2^{n+1} a$ . We have
$2aa_{n+1} + a^2 2^{n+1} \sum_{k=1}^{n} C_{n+1}^{k} = a_{n+1}^2,$
or
$2aa_{n+1} + a^2 2^{n+1} (2^{n+1} - 2) = a_{n+1}^2.$
Since $a_{n+1}$ is positive, we obtain $a_{n+1} = 2^{n+1}a$ , as needed.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.