text: 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.
explanation: The document contains a math problem and its solution. The problem asks to prove that a sequence defined by a specific recurrence relation is constant. The solution is provided, showing the steps to prove the sequence is indeed constant.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: 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.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math problem and its solution. The problem asks to prove that a sequence defined by a specific recurrence relation is constant. The solution is provided, showing the steps to prove the sequence is indeed constant.
  
  Label: question-and-answer