Deriving and Proving the General Formula for the Sequence of Cubes

We know that the general formula for the sequence of odd numbers is 2n - 1 for any natural number n. Furthermore, we can see that the sum of the first n odd integers is equal to n squared. For example,

and so on...........

One can prove this by mathematical induction, but we will not do that in this instance. Well, thinking along these lines I started to think about the sequence of cubes S(1) = 0, 1, 8, 27, 64, 125, ..... Doing the same algorith as before, let's take the difference between each successive term, i.e, a(n) - a(n-1). From this, we get the sequence S(2) = 1, 7, 19, 37, 61, 91,.... When we did this with the sequence of square numbers we got a sequence that was clearly the set of positive odd integers. However, this one is not quite as obvious. So, let's take the difference of each successive term with S(2). Doing this yields S(3) = 6, 12, 18, 24, 30, 36,...... Iterating this process one more time gives us the sequence S(4) = 6, 6, 6, 6,...... So, we can see that the terms in S(3) have a common difference of 6. Also, we see that the terms in S(2) have a common difference of 6n, where n = 1, 2, 3, ..... We will use this information in just a minute.

Again, similar to what we saw with the sequence of squares, the sum of the first n terms of S(2) is equal to n cubed. For example,

and so on.......

However, we cannot prove this by mathematical induction until we can figure out a general formula to represent the sequence 1, 7, 19, 37, 61,.....So, I tried to construct this formula simply using logic. I know from before that the differences between terms in this sequence is 6n, for n = 1, 2, 3, ...., so I used this to help.

By looking at the terms in the sequence I noticed that

1 = 6(0) + 1

7 = 6(1) + 1

19 = 6(3) + 1

37 = 6(6) + 1

61 = 6(10) + 1

and so on.......

From doing this, I realized that the numbers being multiplied by 6 above formed the sequence 0, 1, 3, 6, 10,......, which are the triangular numbers. Well, from previous experience, I know the general formula for triangular numbers is

So, I concluded that my general formula for the sequence 1, 7, 19, 37,.....was

I tried my formula with induction and quickly realized it was not quite perfect, because n = the natural numbers. By that, I mean that a(1) = 7, which is not the first term in my sequence. So, I figured all that I would need to do is subtract back off 6n from each term to get what I needed. This led to what, hopefully, was my completed formula

By simplifying I get:

We can now prove by mathematical induction that

1 + 7 + 19 + 37 + 61 + ..... + (3n^2 - 3n + 1) = n^3

In order to do this, we need to show that the first term is true and the (n + 1) term is true. If we can show those 2 things, then we know that we have found the general formula for our sequence.

STEP 1: Show that our statement is true for n = 1

STEP 2: Assume our statement is true, and show that the n + 1 statement is true

Show:

Now, by substituting what we assumed, we can rewrite our statement as:

Now, just working with the left-hand side we get:

Therefore, by mathematical induction the statement is true! So with this investigation, we have started with nothing and built the general formula for finding n cubed.

return