Cubic Sequences . . . . what do a and b and c and d represent?

Question

I have the following sequence to work out -
1, 7, 25, 63, 129, 231
First line of differences = 6, 18, 38, 66, 102
Second line = 12, 20, 28, 26
Third line = 8, 8, 8

The formula for this is t=an^3 x bn^2 x cn x d
Where t=the total number of cubes in the term

For a quadratic sequence, a = half the second difference (I am assuming that for a cubic sequence, it is half the third difference)
c = 0th term (but how can I work out the 0th term for a cubic sequence, surely d must = the 0th term, but then what does c equal?

And what is d equal to in a cubic sequence?

PS if it helps, I am trying to do 3d borders coursework, and I can't ask my teacher because it's in in the morning!

I've never been taught about gradients. This is UK Year 10 (14/15yo) work!

Answer 1

1) it's actually a 1/6th of the third difference.

If you know anything about differentiation and gradients

n goes up 1 for each consecutive term ("change in x" = 1)
and the change between each t is the "change in y"

"change in y"
------------------ = slope ; but here "change in x" is always 1.
"change in x"

So the "change in y" or more precisely, the change in t, (the differences) relate to gradients at certain points.

Using differentiation:

t=an^3 + bn^2 + cn +d
t'=3an^2 + 2bn + c
t''=3*2an+2b
t'''=3*2a = 6a

Now the gradient or difference in this case is the same anywhere on a straight line

i.e. difference, d = 6a
so a = (d/6).

You may be able to see that equations with x^4 would get differentiated down to

4*3*2*1a

and x^5: 5*4*3*2*1a

so in general for an x^k polynomial (d/k!) = a.


2) d = is the would-be term before the start of your sequence.

This is simply because put in the zeroth term gives d
ie. t(0) = d

c is the zeroth term only when you are dealing with quadratic sequences. This is because c is the number by itself.

Answer 2

The previous answer tells you how to find the a value and so it is 8/6 = 4/3. The easiest way to find the others is to put
n = 1, n = 2 and n = 3 into the formula and equate to the first three numbers in your sequence. Therefore you get
(4/3) + b + c + d = 1
8*(4/3) + 4b + 2c + d = 7
27*(4/3) + 9b + 3c + d = 25
Solving these simultaneous equations is not as difficult as it may seem. Subtracting them in pairs gets rid of the d's and leaves you a pair of simultaneous equations with only two letters. I'm sure that you have learnt how to do those.
In your particular case two of the letters are integers and the other is in thirds like a.