Cubic Sequences

(Difference Method for the $$n^{\text{th}}$$ term)

Cubic sequences are characterized by the fact that the third difference between its terms is constant.

For example, consider the sequence: $4,14,40,88,164, \dots$ looking at the first, second and third difference of this sequence would look like:

Looking at this we can see that the third difference is constant, and not equal to zero, this means it is a cubic sequence.

Tutorial

In the following tutorial we learn how to use the four equations to find the formula for the $$n^{\text{th}}$$ term of a cubic sequence.

Lesson Notes

To read through the lesson notes I give my own students click on the button below.

Formula for the $$n^{\text{th}}$$ term

If a sequence is cubic then its formula can be written: $u_n = an^3+bn^2+cn + d$ For example, the sequence, we saw above: $$4,14,40,88,164, \dots$$ has formula: $u_n = n^3 + 2n^2 - 3n + 4$ Indeed, if we replace $$n$$ by (for example) $$1$$ and $$2$$ we'll find the first and second terms of the sequence, that's: \begin{aligned} u_1 & = 1^3 + 2\times 1^2 - 3 \times 1 + 4 \\ &= 1 + 2 - 3 + 4 \\ u_1 & = 4 \end{aligned}

and
\begin{aligned} u_2 &= 2^3 + 2\times 2^2 - 3 \times 2 + 4 \\ & = 8 + 2\times 4 - 6 + 4 \\ & = 8+8-6+4 \\ u_2 & = 14 \end{aligned} We learn how to find the formula for the $$n^{\text{th}}$$ term below.

Method - Finding the formula for the $$n^{\text{th}}$$ term

Given the first few terms of a cubic sequence, we find its formula $u_n = an^3+bn^2+cn +d$ using the following four equations: $\begin{cases} 6a = \text{third difference} \\ 12a + 2b = \text{1st second difference} \\ 7a + 3b + c = \text{difference between the first two terms} \\ a + b + c + d = \text{first term} \end{cases}$ This provides us with four equations, which we solve from top to bottom, to find each of the four unknown coefficients $$a$$, $$b$$, $$c$$ and $$d$$.

The four equations, stated above, can be a little confusing so we illustrate it here here with an example.

Consider the sequence whose first few terms are: $4,14,40,88,164, \dots$ The following illustration shows where each of the four equations, stated above, fits in.:

We can see that the values that we use, the ones we've boxed in the illustration, are always the first on each row.

We therefore have four equations to solve: $\begin{cases} 6a = 6 \\ 12a + 2b = 16 \\ 7a + 3b + c = 10 \\ a + b + c + d = 4 \end{cases}$ We solve these working from top to bottom. Each time we move-on to solve a new equation we substitute the value we just found in the previous equations:

• equation 1:
We start with the first (top) equation: \begin{aligned} & 6a = 6 \\ & a = \frac{6}{6} \\ & a = 1 \end{aligned} Now that we know the value of $$a$$, $$a =1$$, we move-on to the second equation.
• equation 2:
We have to solve $$12a + 2b = 16$$. Now that we know that $$a=1$$ this becomes: \begin{aligned} & 12\times 1 + 2b = 16 \\ & 12 + 2b = 16 \\ & 2b = 16 - 12 \\ & 2b = 4 \\ & b = \frac{4}{2} \\ & b = 2 \end{aligned} Now that we know the value of $$b$$, $$b =2$$, we move-on to the third equation.
• equation 3:
We have to solve $$7a + 3b + c = 10$$. Since we know that $$a=1$$ and $$b = 2$$ this becomes: \begin{aligned} & 7 \times 1 + 3 \times 2 + c = 10 \\ & 7 + 6 + c = 10 \\ & 13 + c = 10 \\ & c = 10 - 13 \\ & c = -3 \end{aligned} Now that we know the value of $$c$$, $$c = -3$$, we move-on to the fourth equation.
• equation 4:
We have to solve $$a + b + c + d = 4$$. Since we know that $$a=1$$, $$b = 2$$ and $$c = -3$$ this becomes: \begin{aligned} & 1 + 2 - 3 + d = 4 \\ & 3 - 3 + d= 4 \\ & d = 4 \end{aligned} Now that we know the value of $$d$$, $$c = 4$$, we can finally state the formula for the n-th term of this cubic sequence: $u_n = n^3 + 2n^2 - 3n + 4$

Exercise

Find the formula for the $$n^{\text{th}}$$ term of each of the following sequences:

1. The sequence whose first few terms are: $4,14,40,88,164, \dots$
2. The sequence whose first few terms are: $4,23,66,145,272, \dots$
3. The sequence whose first few terms are: $-1,1,-5,-25,-65, \dots$
4. The sequence whose first few terms are: $1,14,65,178,377, \dots$
5. The sequence whose first few terms are: $11,6,-25,-100,-237, \dots$

Note: this exercise can be downloaded as a worksheet to practice with:

1. $$u_n = n^3+2n^2-3n+4$$

2. $$u_n = 2n^3+5n-3$$

3. $$u_n = -n^3+2n^2+3n-5$$

4. $$u_n = 4n^3-5n^2+2$$

5. $$u_n = -3n^3+5n^2+n+8$$