Question 451744: Find the pattern with these numbers: 2. 9. 27. 65
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Here is a possible answer. I can't guarantee it is the only one.
2 to 9, difference 7
9 to 27, difference 18
27 to 65, difference 38
Second tier differences:
7 to 18, difference 11
18 to 38, difference 20
Third tier:
11 to 20, difference 9
Here is where the analysis gets a little dicey. Generally you would keep taking sub-tier differences until you found that they were all the same, and then use the number of tiers to determine the degree of the polynomial required to model the sequence. Here you got down to where there is only one difference, so you don't know whether it is the same as anything or not. It is certainly true that a 3rd degree polynomial will model this sequence (which we will find momentarily), but what is not known is whether or not the constructor of the problem intended a different sequence with the same 4 first terms that is modeled by a polynomial of greater degree. In fact there are different sequences, an infinity of them, hence my caveat at the beginning of this solution.
Assume that we are satisfied with the sequence modeled by a 3rd degree polynomial of the form:
Where  is the ordinal number of the term
So for term #1 which has a value 2,
^3\ +\ b(1)^2\ +\ c(1)\ +\ d\ =\ 2)
Or
Then for term #2 which has a value of 9:
^3\ +\ b(2)^2\ +\ c(2)\ +\ d\ =\ 9)
Or
Likewise we derive:
and
Giving a 4 by 4 linear system. You can use any of several methods to solve. I use the MDETERM function in Excel and then apply Cramer's Rule. By doing that, I obtained:
and therefore the polynomial function for the  term as a function of  is:
\ =\ \frac{3}{2}n^3\ -\ \frac{7}{2}n^2\ +\ 7n\ -\ 3)
To check, go back to the differences analysis. Add 9 to 20 to get 29. Then add 29 to 38 to get 67. Finally add 67 to 65 to get 132.
Then I'll leave it to you to do the arithmetic to evaluate and verify that
\ =\ 132)
John
My calculator said it, I believe it, that settles it
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