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Math Help - General Formula For Sequence

  1. #1
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    General Formula For Sequence

    Find the general formula (for the nth term) for the sequence

    {5,1,5,1,5,1,...}

    assuming the pattern continues.

    I've tried a lot of things but can't come up with a solution...it seems innocent enough. It seems like the formula should depend on n being even or odd but I can't get anything to work. A suggestion would be appreciated. Is there a specific way to solve these or are you supposed to just spot the pattern. Thanks.
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  2. #2
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    Quote Originally Posted by vReaction View Post
    Find the general formula (for the nth term) for the sequence

    {5,1,5,1,5,1,...}

    assuming the pattern continues.

    I've tried a lot of things but can't come up with a solution...it seems innocent enough. It seems like the formula should depend on n being even or odd but I can't get anything to work. A suggestion would be appreciated. Is there a specific way to solve these or are you supposed to just spot the pattern. Thanks.


    a_n=\left\{\begin{array}{ll}5&,\,if\,\,\,n\,\,\,is  \,\,\,odd\\1&,\,if\,\,\,n\,\,\,is\,\,\,even\end{ar  ray}\right.

    Tonio
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  3. #3
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    Quote Originally Posted by vReaction View Post
    Find the general formula (for the nth term) for the sequence {5,1,5,1,5,1,...}, assuming the pattern continues.
    Try x_n=5\cdot \text{mod}(n,2)+\text{mod}(n+1,2).
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  4. #4
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    Tonio: Thanks...I didn't think it'd be that simple.

    Plato: I'm unfamiliar with "mod"...I searched and only found modular arithmetic which looked very confusing. Thanks though.
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  5. #5
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    Quote Originally Posted by vReaction View Post
    : I'm unfamiliar with "mod"...I searched and only found modular arithmetic which looked very confusing.
    It is not confusing at all.
    \text{mod}(j,k) is the remainder when j is divided by k.
    So \text{mod}(3,2)=1 and \text{mod}(4,2)=0

    \text{mod}(5,3)=2 and \text{mod}(12,6)=0
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  6. #6
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    Hello, vReaction!

    Find the n^{th} term for the sequence: . 5,1,5,1,5,1\:\hdots

    The sequence is: . (3+2),\;(3-2),\;(3+2),\;(3-2),\;\hdots


    Therefore: . a_n \;=\;3 - (\text{-}1)^n\!\cdot\!2


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    The "Blinker" function


    Consider these two functions:

    . . \frac{1-(\text{-}1)^n}{2} \;=\;\begin{Bmatrix}1 & n \text{ odd} \\ 0 & n\text{ even} \end{Bmatrix}

    . . \frac{1+(\text{-}1)^n}{2} \;=\;\begin{Bmatrix} 0 & n\text{ odd} \\ 1 & n\text{ even} \end{Bmatrix}



    Given the sequence: . A,B,A,B,A,B\:\hdots

    . . then: . a_n \;\;=\;\;\frac{1-(\text{-}1)^n}{2}\cdot A \:+\: \frac{1+(\text{-}1)^n}{2}\cdot B


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