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Math Help - Algebraic definition for the nth term

  1. #1
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    Algebraic definition for the nth term

    Determine the next two numbers in the following sequence
    and determine an explicit algebraic definition for the nth term.

    43, 44, 50, 53, 65, 60, 88, 95, 119,
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  2. #2
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    are you sure its not
    43,44,50,53,65,70,88,95,119 ?

    the differences between terms follows the pattern

    1,6,3,12,5,18,7,24 which means the differnce in the next 2 terms would be 9,30 giving 128 and 158 as the next 2 terms

    the terms alternate increasing by 2n+1 and 6n
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  3. #3
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    No, it's 43, 44, 50, 53, 65, 60, 88, 95, 119,
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  4. #4
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    But I sincerely feel that it was a mistake, because I found the same pattern, but was disgruntled when approaching the 60, so I think that it has to be 70.
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  5. #5
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    If this is so an it really is 70, then what would be the explicit algebraic definition for the nth term? And how do you do this?
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  6. #6
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    Hello, warriors8371

    If the sixth term is 70, there is an interesting solution.


    Determine the next two numbers in the following sequence
    and determine an explicit algebraic definition for the n^{th} term.

    . . <br />
    43, 44, 50, 53, 65, 70, 88, 95, 119, \hdots

    Consider the two subsequences formed by odd n and even n.


    \begin{array}{c|ccccc}<br />
\text{odd }n & 1 & 3 & 5 & 7 & 9 \\ \hline<br />
a_n & 43 & 50 & 65 & 88 & 119 \end{array}

    . . The generating function is: . a_n \;=\;\frac{2n^2 - n +85}{2} .[1] .for odd n.


    \begin{array}{c|cccc}<br />
\text{even }n & 2 & 4 & 6 & 8 \\ \hline<br />
a_n & 44 & 53 & 70 & 95 \end{array}

    . . The generating function is: . a_n \;=\;\frac{2n^2 - 3n + 86}{2} .[2] .for even n.


    To make the two functions alternate:

    . . (1) Multiply [1] by \frac{1 - (-1)^n}{2}

    . . (2) Multiply [2] by \frac{1 + (-1)^n}{2}

    . . (3) Add the two results.


    We have: . a_n \;=\;\frac{1-(-1)^n}{2}\left(\frac{2n^2-n+85}{2}\right) + \frac{1+(-1)^n}{2}\left(\frac{2n^2 - 3n + 86}{2}\right)

    . . which simplifies to: . \boxed{a_n \;=\;\frac{4n^2 - 4n + 171 + (-1)^n(1-2n)}{4}}


    The next two terms are:

    . . a_{10} \;=\;\frac{4(10^2) - 4(10) + 171 + (-1)^{10}(1-20)}{4} \;=\;\frac{400 - 40 + 171 - 19}{4} \;= . \frac{512}{4} \;=\;\boxed{128}

    . . a_{11} \;=\;\frac{4(11^2) - 4(11) + 171 + (-1)^{11}(1-22)}{4} \;=\;\frac{484 - 44 + 171 + 21}{4} \;= . \frac{632}{4} \;=\;\boxed{158}

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