Expert: Josh - 11/16/2006

Question
i have been trying to figure this sequence out for ages could you help?

find the formula for the nth term of:

-4,2,26,74,152,266

Answer
Hi Harry,

In short, I do not know what function would fit this sequence. In the time that I have, I can definitely rule out several possibilities, to the extent that if it fits any polynomial function at all, it would at least have order 3. Here is what I have tried (unsuccessfully).

First of all, this is neither an arithmetic, nor geometric sequence. i.e., each successive term is not related to the previous term by addition or multiplication of a constant factor.

Since the sequence is increasing, we should consider the possibility of a polynomial function, with exponent (or degree) greater than 1. It also helps if you try plotting the sequence on a piece of paper to get a rough idea what it is that we are looking at. From such a graph, one would guess that the sequence may fit some sort of parabola (i.e., a polynomial of degree two).

x= 1   2   3   4   5
y=-4   2  26  74  152

Parabola with shape "c", symmetrical about x=m, with
vertical offset "k" is described by the equation,
y(x)=c*(x-m)^2+k
Observe that this expression contains 3 parameters
or unknowns. The value of "c" is positive as our
sketch shows that the curve concaves up (y values
increase with x). The parameter "c" basically controls
how 'fat' or 'thin' the parabola looks.

Assuming the sequence fits a parabola:
When x=1, -4=c(1-m)^2+k ...[1]
When x=2,  2=c(2-m)^2+k ...[2]
When x=3, 26=c(3-m)^2+k ...[3]

Note: we need 3 equations to solve for 3 unknowns.
Rearrange [1], making k the subject
k=-(4+c(1-m)^2) ...[4]
Substitute for k in expressions [2] and [3],
[2] =>  2=c(2-m)^2-(4+c(1-m)^2)
[3] => 26=c(3-m)^2-(4+c(1-m)^2)
Expand carefully,
[2] =>  2=c(4-4m+m^2)-(4+c(1-2m+m^2))
[3] => 26=c(9-6m+m^2)-(4+c(1-2m+m^2))
Note: now we deal with 2 equations and 2 unknowns.

Simplify expression by gathering like terms,
For expression [2],
2=c(4-4m+m^2)-(4+c(1-2m+m^2))
6=c(4-4m+m^2-(1-2m+m^2))
6=c(3-2m)
c=6/(3-2m) ...call this [5]

For expression [3],
26=c(9-6m+m^2)-(4+c(1-2m+m^2))
30=c(9-6m+m^2-(1-2m+m^2))
30=c(8-4m) ...call this [6]

Finally, put [5] into [6],
30=6(8-4m)/(3-2m)
5(3-2m)=(8-4m)
15-10m=8-4m
7=6m
m=7/6 ...we obtain the "symmetry parameter"
This shows that the sequence fits a parabola
which is symmetrical about x=7/6.

Plug this back into [5], we obtain
c=6/(3-2*(7/6))
c=6/(3-7/3)
c=6/((9-7)/3)
c=18/2
c=9
This is what we expected, a positive value as
explained earlier.

Put m=7/6 and c=9 into [4],
k=-(4+c(1-m)^2)
This evaluates to k=-17/4, i.e., when x=0, the parabola
formed by the sequence cuts the y-axis at y=-17/4.

Thus, the sequence fits the parabolic equation:
y(x) = c*(x-m)^2+k = 9*(x-(7/6))^2-(17/4)
In your question, you can treat "x" as the same as "n"

If our assumption holds, then the general expression for
the nth term in the sequence would be y(n) = 9*(n-(7/6))^2-(17/4).

Unfortunately, it does not agree with the sequence beyond the terms we have tried. Thus, we can conclude that it is more complex than a parabola. If you are familiar with least square regression techniques, you can perhaps try that. But that goes way beyond the scope of high school maths. Are you in college if you don't mind me asking?