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Math Help - Sequences/Patters

  1. #1
    Senior Member euclid2's Avatar
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    Sequences/Patters

    given the patterns
    a) 4,8,12,16,20,24(going up by 4)
    b) 3,9,27,81,243,729(* by 3)
    c) 144,121,100,81,64,49(-21,-21,-19...)
    d) 3,-12,48,-192,728,-3072(*-4)

    Find an expression for the general term for the sequences listed above(one general term for each sequence)
    Can you please explain how to do it aswell
    thanks
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  2. #2
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    [quote=euclid2;186568]given the patterns
    a) 4,8,12,16,20,24(going up by 4)

    4,8,12,16,20............ \hfill \\

     {\text{This arithmetic sequence has first term, }}a = 4{\text{ and common diffrence, }}d = 8 - 4 = 4 \hfill \\

     {\text{General term, }}t_n  = a + \left( {n - 1} \right)d \hfill \\

     = 4 + \left( {n - 1} \right)4 \hfill \\

      = 4 + 4n - 4 = 4n \hfill \\

      t_n  = 4n \hfill \\
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    [quote=euclid2;186568]given the patterns

    b) 3,9,27,81,243,729(* by 3)

      3,9,27,81,243,................... \hfill \\

    {\text{This geometric sequence has first term, }}a = 3{\text{ and common ratio, }}r = \frac{9}<br />
{3} = 3 \hfill \\

      {\text{General term, }}t_n  = ar^{n - 1}  \hfill \\

      = 3\left( 3 \right)^{n - 1}  \hfill \\

     = 3^{1 + n - 1}  = 3^n  \hfill \\

      t_n  = 3^n  \hfill \\
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    [quote=euclid2;186568]given the patterns

    d) 3,-12,48,-192,728,-3072(*-4)

      3, - 12,48, - 192,728, - 3072,................... \hfill \\

    {\text{This geometric sequence has first term, }}a = 3{\text{ and common ratio, }}r = \frac{{ - 12}}<br />
{3} =  - 4 \hfill \\

    {\text{General term, }}t_n = ar^{n - 1} \hfill \\

     = 3\left( { - 4} \right)^{n - 1}  \hfill \\

    t_n  = 3\left( { - 4} \right)^{n - 1}  \hfill \\
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  5. #5
    Senior Member euclid2's Avatar
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    thank you very much that is extremely helpful
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    [quote=euclid2;186568]given the patterns

    c) 144,121,100,81,64,49(-21,-21,-19...)

     144,121,100,81,64,49,................. \hfill \\

     12^2, 11^2, 10^2, 9^2,...............

     {\text{This sequence is neither arithmetic nor geometric}} \hfill \\

     \text { the sequence is}\;\; (13-1)^2, (13-2)^2, (13-3)^2,......

    t_n=(13-n)^2
    Last edited by Shyam; September 13th 2008 at 02:19 PM.
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  7. #7
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    Hello, euclid2!


    Given the patterns:

    a) 4, 8, 12, 16, 20, 24, ...

    b) 3, 9, 27, 81, 243, 729, ...

    c) 144, 121, 100, 81, 64, 49, ...

    d) 3, -12, 48, -192, 728, -3072, ...

    Find an expression for the general term for the sequences.

    We must examine the pattern and create a function in terms of n.


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


    a)\quad\begin{array}{c|c|c|c|c|c|c| c}<br />
n & {\color{red}1} & {\color{red}2} & {\color{red}3} & {\color{red}4} & {\color{red}5} & {\color{red}6} & \hdots \\ \hline<br />
a_n & 4 & 8 & 12 & 16 & 20 & 24 & \hdots \\<br />
& 4\cdot{\color{red}1} & 4\cdot{\color{red}2} & 4\cdot{\color{red}3} & 4\cdot{\color{red}4} & 4\cdot{\color{red}5} & 4\cdot{\color{red}6} & \hdots \\ \hline \end{array}

    We see that a_n is always: \text{4 times }n.

    Therefore: . a_n \;=\;4n


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


    b)\quad\begin{array}{c|c|c|c|c|c|c|c}<br />
n & {\color{red}1} & {\color{red}2} & {\color{red}3} & {\color{red}4} & {\color{red}5} & {\color{red}6} & \hdots \\ \hline<br />
a_n & 3 & 9 & 27 & 81 & 243 & 729 & \hdots \\<br />
& 3^{{\color{red}1}} & 3^{{\color{red}2}} & 3^{{\color{red}3}} & 3^{{\color{red}4}} & 3^{{\color{red}5}} & 3^{{\color{red}6}} & \hdots \\ \hline<br />
\end{array}

    We see that a_n is always: \text{3 to the }n^{th}\text{ power.}

    Therefore: . a_n \;=\;3^n


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


    c)\quad\begin{array}{c|c|c|c|c|c|c|c}<br />
n & {\color{red}1} & {\color{red}2} & {\color{red}3} & {\color{red}4} & {\color{red}5} & {\color{red}6} & \hdots \\ \hline<br />
a_n & 144 & 121 & 100 & 81 & 64 & 49 & \hdots \\<br />
& 12^2 & 11^2 & 10^2 & 9^2 & 8^2 & 7^2 & \hdots \\<br />
& (13-{\color{red}1})^2 & (13-{\color{red}2})^2 & (13-{\color{red}3})^2 & (13-{\color{red}4})^2 & (13-{\color{red}5})^2 & (13-{\color{red}6})^2 & \hdots \end{array}

    We see that a_n is always: \text{the square of }(13-n)

    Therefore: . a_n \;=\;(13-n)^2


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


    d)\quad\begin{array}{c|c|c|c|c|c|c|c}<br />
n & {\color{red}1} & {\color{red}2} & {\color{red}3} & {\color{red}4} & {\color{red}5} & {\color{red}6} & \hdots \\ \hline<br />
a_n & 3 & \text{-}12 & 48 & \text{-}192 & 728 & \text{-}3072 & \hdots \\<br />
& 3(\text{-}4)^{{\color{red}0}} & 3(\text{-}4)^{{\color{red}1}} & 3(\text{-}4)^{{\color{red}2}} & 3(\text{-}4)^{{\color{red}3}} & 3(\text{-}4)^{{\color{red}4}} & 3(\text{-}4)^{{\color{red}5}} & \hdots\end{array}

    We see that a_n is always: \text{3 times a power of -4,}
    . . \text{and that power is }n-1.

    Therefore: . a_n \;=\;3\cdot(\text{-}4)^{n-1}

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