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Thread: sequence problem

  1. #1
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    sequence problem

    given the sequence :

    $\displaystyle x_{n+1} =\frac{6x_{n}+1}{2x_{n}+5}$

    find its general term
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  2. #2
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    Quote Originally Posted by xalk View Post
    given the sequence :

    $\displaystyle x_{n+1} =\frac{6x_{n}+1}{2x_{n}+5}$

    find its general term
    Hi

    If $\displaystyle x_0 = -\frac12$ then $\displaystyle \forall n~
    x_n = -\frac12$

    If $\displaystyle x_0 \neq -\frac12$ then let $\displaystyle y_n = \frac{x_n - 1}{2\:x_n + 1}$

    You can show that $\displaystyle y_{n+1} = \frac47\:y_n$ leading to $\displaystyle y_n = \alpha\:\left(\frac47\right)^n$ where $\displaystyle \alpha = y_0 = \frac{x_0 - 1}{2\:x_0 + 1}$

    Spoiler:

    And finally $\displaystyle x_n = \frac{7^n + \alpha\:4^n}{7^n - 2 \alpha\:4^n}$

    When $\displaystyle \alpha \rightarrow \pm\infty$ $\displaystyle x_n \rightarrow -\frac12$
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  3. #3
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    Quote Originally Posted by running-gag View Post
    Hi

    If $\displaystyle x_0 = -\frac12$ then $\displaystyle \forall n~
    x_n = -\frac12$

    If $\displaystyle x_0 \neq -\frac12$ then let $\displaystyle y_n = \frac{x_n - 1}{2\:x_n + 1}$

    You can show that $\displaystyle y_{n+1} = \frac47\:y_n$ leading to $\displaystyle y_n = \alpha\:\left(\frac47\right)^n$ where $\displaystyle \alpha = y_0 = \frac{x_0 - 1}{2\:x_0 + 1}$

    Spoiler:

    And finally $\displaystyle x_n = \frac{7^n + \alpha\:4^n}{7^n - 2 \alpha\:4^n}$

    When $\displaystyle \alpha \rightarrow \pm\infty$ $\displaystyle x_n \rightarrow -\frac12$
    THANK you,suppose the 1st member is 3 what is the limit of the sequence.

    I am a Little confused
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  4. #4
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    Do you mean $\displaystyle x_0 = 3$ ?

    Then $\displaystyle \alpha = \frac{3 - 1}{2\cdot3 + 1} = \frac27$

    $\displaystyle x_n = \frac{7^n + \alpha\:4^n}{7^n - 2 \alpha\:4^n} = \frac{1 + \alpha\:\left(\frac47\right)^n}{1 - 2 \alpha\:\left(\frac47\right)^n}$

    Therefore $\displaystyle \lim_{n\rightarrow +\infty} x_n = 1$
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