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nth term formulas
superres Oct 26, 2005 12:00 AM (in response to butterdisabled)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 nonzero 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*6^{2}+5
and the general form is:
6*n^{2}+(n1) where n starts at 1
or
6*(n+1)^{2}+n where n starts at 0
TTFN,
Eden 
nth term formulas
butterdisabled Oct 26, 2005 12:00 AM (in response to butterdisabled)TO TOM:
I can figure out the next numbers, what is the nth term equation e.g.: 2n(n3) that will get me the correct next number??
nth term formulas
TomGutman Oct 26, 2005 12:00 AM (in response to butterdisabled)See my reply to your other thread. You can plug in the differences to the general formula and get the expression 6n²+n1. You can factor that into (2n+1)(3n1), although I don't see that as much of an improvement.
� � � � Tom Gutman 
nth term formulas
superres Oct 26, 2005 12:00 AM (in response to butterdisabled)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(n3) that
>will get me the correct next
>number??
the second difference of 12 tells you that the n^{2} term is multiplied by 6. It's essentially the derivative of the power term in the equation. If you then subtract 6x^{2} 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


nth term formulas
TomGutman Oct 26, 2005 12:00 AM (in response to butterdisabled)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
nth term formulas
butterdisabled Oct 26, 2005 12:00 AM (in response to TomGutman)On 10/26/2005 4:39:39 PM, Tom_Gutman wrote:
>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 
Re: nth term formulas
Liv Oct 21, 2010 11:15 AM (in response to TomGutman)Nice, like Babbage machine ...
Best, Liv


nth term formulas
TomGutman Oct 26, 2005 12:00 AM (in response to butterdisabled)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 
nth term formulas
butterdisabled Oct 26, 2005 12:00 AM (in response to butterdisabled)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
nth term formulas
superres Oct 26, 2005 12:00 AM (in response to butterdisabled)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 n^{2}
The first differences are 3,5,7,9,11 and the second differences are 2,2,2,2.
say the series is 2n^{2}
The first differences are 6,10,14,18,22 and the second differences are 4,4,4,4.
TTFN,
Eden
