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Math Help - sequence problem

  1. #1
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    sequence problem

    given the sequence :

    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 :

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

    find its general term
    Hi

    If x_0 = -\frac12 then \forall n~<br />
x_n = -\frac12

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

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

    Spoiler:

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

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

    If x_0 = -\frac12 then \forall n~<br />
x_n = -\frac12

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

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

    Spoiler:

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

    When \alpha \rightarrow \pm\infty 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 x_0 = 3 ?

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

    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 \lim_{n\rightarrow +\infty} x_n = 1
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