Home > Analysis, Probability, Problem Solving > Recurrent Sequence implies Constant

## Recurrent Sequence implies Constant

Prove that the sequence which satisfies the relation $a_{n+1}=\sqrt{6-2a_n^2},\ \forall n\geq 1$ is constant.
Laurentiu Panaitopol, Romania
Solution: We can translate the recurrence relation in the following way:
$a_{n+1}^2+2a_n^2=6,\ (0)$. Therefore $a_{n+2}^2+2a_{n+1}^2=6=a_{n+1}^2+2a_n^2$, which implies $a_{n+2}^2-a_{n+1}^2=(-2)(a_{n+1}^2-a_n^2),\ (1)$.
From $(0)$ it follows that $(a_n)$ is bounded, and from $(1)$ it follows that $y_n=a_{n+1}^2-a_n^2$ is unbounded if it is not zero. This would contradict the fact that $(a_n)$ is bounded. Therefore $(y_n)$ is bounded and $(a_n^2)$ is constant. Because $a_n \geq 0,\ \forall n \geq 2$ we get that $(a_n)$ is constant.
The text passage contains a mathematical problem and its solution, but it does not contain a direct question or an answer in the form of a query and response.
no_question
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
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- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
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Home > Analysis, Probability, Problem Solving > Recurrent Sequence implies Constant

## Recurrent Sequence implies Constant

Prove that the sequence which satisfies the relation $a_{n+1}=\sqrt{6-2a_n^2},\ \forall n\geq 1$ is constant.
Laurentiu Panaitopol, Romania
Solution: We can translate the recurrence relation in the following way:
$a_{n+1}^2+2a_n^2=6,\ (0)$. Therefore $a_{n+2}^2+2a_{n+1}^2=6=a_{n+1}^2+2a_n^2$, which implies $a_{n+2}^2-a_{n+1}^2=(-2)(a_{n+1}^2-a_n^2),\ (1)$.
From $(0)$ it follows that $(a_n)$ is bounded, and from $(1)$ it follows that $y_n=a_{n+1}^2-a_n^2$ is unbounded if it is not zero. This would contradict the fact that $(a_n)$ is bounded. Therefore $(y_n)$ is bounded and $(a_n^2)$ is constant. Because $a_n \geq 0,\ \forall n \geq 2$ we get that $(a_n)$ is constant.
user
The text passage contains a mathematical problem and its solution, but it does not contain a direct question or an answer in the form of a query and response.

TYPE: NO_QUESTION