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    The question is:

    Find the nth term of the following sequence:

    5, 7, 13, 23, 37
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    (Original post by lalallla22)
    The question is:

    Find the nth term of the following sequence:

    5, 7, 13, 23, 37
    The first differences are:

    2, 6, 10, 14

    There is a constant second difference of 4 which means this is a quadratic sequence. Have you learnt how to find the nth term of a quadratic sequence? If you have, please post what you've tried.
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    [QUOTE=Notnek;74130646]The first differences are:

    2, 6, 10, 14


    the common difference is 4 which you then half and multiply by n^2

    you then right out the 2n^2 sequence as well as the n^2 sequence


    2n^2 = 2 8 18 32 50
    n^2 = 1 4 9 16 25


    The next part I believe I have done incorrectly, however, I shall continue.


    I took the 2n^2 sequence away from the original sequence

    5 | 7 | 13 | 23 | 37|
    - | | | |
    2 | 8 | 18 | 32 |50 |
    ________________

    3 | -1| -5 | -9 |-13 |
    ________________


    Then I found the nth term for the sequence above

    3, -1, -5, -9, -13
    -4 -4 -4 -4

    -4n + 7



    Then I added it to the first part of the nth term for the initial sequence.

    2n^2 - 4n + 7


    However, if I then apply that to n:


    2* 1^2 - 4*1 + 7

    = 2 - 4 +7
    = 5


    2* 2^2 - 4*2 + 7

    = 8 - 8 + 7
    = 7

    2* 3^2 - 4*3 + 7

    =18 - 12 + 7
    = 13


    Oh, it seems that I have solved it. Never mind.
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    • the sequence can be written as:5, 5 2,5 2 6 ...
    • so it is 5 plus the sum of the sequence 2,6,10,14
    • use the sum of geometric series to find how much to add to 5
    • sn= n/2(2a (n-1)d)
    • n=n-1 as the first tem of the sequence 5,7,13 has none of the other sequence added on
    • a=2
    • d=4
    • therefore the sum of the series is 5 + (n-1)/2 (2(2) (n-2)4)
    • which can be simplified to 5 + 2(n-1)^2
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    In case it's any help to anyone, here's an example of finding the nth term of a quadratic sequence, using my favourite method.
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