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3167 Views 10 Replies Latest reply: Oct 21, 2010 11:15 AM by Liv RSS
Copper 5 posts since
Oct 26, 2005
Currently Being Moderated

Oct 26, 2005 12:00 AM

nth term formulas

I am trying to help my daughter in her Geometry I class. Is there a formula for finding the nth term in a sequence? My oldest daughter, BA degree, and myself have tried to figure out the nth term. One problem looks like this:

Find the nth term of

6, 25, 56, 99, 154

We have NO idea on how to start on this one.
Any help will be appreciated. Thank you
  • superres Silver 7,759 posts since
    Jun 1, 2007
    Currently Being Moderated
    Oct 26, 2005 12:00 AM (in response to butter-disabled)
    nth term formulas
    On 10/26/2005 1:27:19 PM, butter wrote:
    >I am trying to help my
    >daughter in her Geometry I
    >class. Is there a formula for
    >finding the nth term in a
    >sequence? My oldest daughter,
    >BA degree, and myself have
    >tried to figure out the nth
    >term. One problem looks like
    >this:
    >
    >Find the nth term of
    >
    > 6, 25, 56, 99, 154
    >
    >We have NO idea on how to
    >start on this one.
    >Any help will be appreciated.
    >Thank you

    Graphing such things usually helps. I did it in Excel. If you do that, you'll see that it's almost perfect fit for a series of squares of the first 5 non-zero integers multiplied by a scaling factor of 6. So, if you subtract that from the original numbers, you get the first 5 integers, e.g., those guys start at 0.

    Therefore, the next number should be 6*62+5
    and the general form is:

    6*n2+(n-1) where n starts at 1
    or
    6*(n+1)2+n where n starts at 0

    TTFN,
    Eden
    • TomGutman Silver 10,537 posts since
      Oct 22, 2006
      Currently Being Moderated
      Oct 26, 2005 12:00 AM (in response to butter-disabled)
      nth term formulas
      See my reply to your other thread. You can plug in the differences to the general formula and get the expression 6n²+n-1. You can factor that into (2n+1)(3n-1), although I don't see that as much of an improvement.

      � � � � Tom Gutman
    • superres Silver 7,759 posts since
      Jun 1, 2007
      Currently Being Moderated
      Oct 26, 2005 12:00 AM (in response to butter-disabled)
      nth term formulas
      On 10/26/2005 5:13:31 PM, butter wrote:
      >TO TOM:
      >
      >I can figure out the next
      >numbers, what is the nth term
      >equation e.g.: 2n(n-3) that
      >will get me the correct next
      >number??

      the second difference of 12 tells you that the n2 term is multiplied by 6. It's essentially the derivative of the power term in the equation. If you then subtract 6x2 from each term, the resultant is linear term series whose 1st difference is 1, so there is a 1*n term. subtract that from the original series leaves the constant term.

      TTFN,
      Eden
  • TomGutman Silver 10,537 posts since
    Oct 22, 2006
    Currently Being Moderated
    Oct 26, 2005 12:00 AM (in response to butter-disabled)
    nth term formulas
    There is no fully general method. Much depends on the particular series. But a useful place to start is to calculate the differences:

    6, 25, 56, 99, 154

    19, 31, 43, 55

    12, 12, 12

    The second differences are constant, so the numbers can be represented by a quadratic equation. But there is no need to actually work out the quadratic -- the table of differences can be extended by adding a fourth second difference of 12:

    6, 25, 56, 99, 154, 221

    19, 31, 43, 55, 67

    12, 12, 12, 12


    � � � � Tom Gutman
    • Liv Bronze 41 posts since
      Oct 4, 2009
      Currently Being Moderated
      Oct 21, 2010 11:15 AM (in response to TomGutman)
      Re: nth term formulas

      Nice, like Babbage machine ...

       

      Best, Liv

  • TomGutman Silver 10,537 posts since
    Oct 22, 2006
    Currently Being Moderated
    Oct 26, 2005 12:00 AM (in response to butter-disabled)
    nth term formulas
    You're mixing up the threads. This sequence is the one from the other thread, which I covered in detail there. What tells you that the equation is a quadratic is not the value of the second differences, but the fact that the second differences are all equal. In the two series that you have the second differences are 12 in one and 4 in the other. But both have a quadratic form, because it is the second differences that are constant. If it were the third differences that were constant the form would be a cubic.

    � � � � Tom Gutman
    • superres Silver 7,759 posts since
      Jun 1, 2007
      Currently Being Moderated
      Oct 26, 2005 12:00 AM (in response to butter-disabled)
      nth term formulas
      On 10/26/2005 5:57:23 PM, butter wrote:
      >So if I have these terms:
      >
      > -3 0 7 18 33 52
      >3 7 11 15 19
      >differences
      >4 4 4 4
      >differences
      >
      >the second row of differences
      >is constant (4) so my formula
      >should be n+4 ??????..
      >
      >I understand how to get the
      >differences but when you say
      >that by looking at a constant
      >(4) the polynomial that
      >generates the sequence will be
      >a degree of ????....this is
      >where I'm stuck.
      >
      >HELP

      The second difference being a constant tells you that the equation is quadratic.

      say the series is n2
      The first differences are 3,5,7,9,11 and the second differences are 2,2,2,2.

      say the series is 2n2
      The first differences are 6,10,14,18,22 and the second differences are 4,4,4,4.


      TTFN,
      Eden

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