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    Q: Find the Nth term: 3 9 18 30 45

    What I've done: 1st diff = 6 9 12 15....

    2nd diff = 2

    so the nth term involves n^2



    dunno what to do next? what is an expression for the nth term?

    cheeeers xxxxxxxxxxxx
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    n^2 = 1 4 9 16 25

    is that relevant?
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    That sequence is just 3x the triangle numbers
    Spoiler:
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    \dfrac{3n}{2}(n+1)
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    the 2nd difference is 3 not 2
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    (Original post by Potassium^2)
    That sequence is just 3x the triangle numbers
    Spoiler:
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    \dfrac{3n}{2}(n+1)
    nice one!
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    If you get one of these where you don't spot a nice pattern (like it being a multiple of the triangle numbers), say you notice that the second difference is constant. Then you can let the nth term be equal to an^2+bn+c, and then you can substitute in three values of n to find out what a,b,c are. It's a bit of a pain to do though, so spotting a pattern is usually a good idea!
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    (Original post by nuodai)
    If you get one of these where you don't spot a nice pattern (like it being a multiple of the triangle numbers), say you notice that the second difference is constant. Then you can let the nth term be equal to an^2+bn+c, and then you can substitute in three values of n to find out what a,b,c are. It's a bit of a pain to do though, so spotting a pattern is usually a good idea!
    It's also worth knowing that,

    if it's a quadratic sequence then the coefficicent of n^2 = second difference / 2!

    and

    if it's a cubic sequence then the coefficicent of n^3 = third difference / 3!

    and so on.

    You can also just write down the answer.

    e.g. 3, 9, 18, 30, 45

    3\frac{(n-2)(n-3)}{(1-2)(1-3)}+9\frac{(n-1)(n-3)}{(2-1)(2-3)}+18\frac{(n-1)(n-2)}{(3-1)(3-2)}
 
 
 
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