articles:sequences

Given the integer (whole number) sequence

3, 6, 9, 12, …

what is the next number in the sequence?

The sequence of numbers in this question you should recognise as the 3 times table. The next number in the sequence will be 15 ( $3 \times 5$ )

In this sequence the first term is 3, the second is 6, the third is 9, and so on. This leads us to the next question: can you write down an expression for the n^th term in this sequence?

The answer to this is as this is the 3 times table then the value of the n^th term will be

$3 \times n$ , or $3n$

Unfortunately, questions on sequences will never be this simple. But this article gives you hints and tips on how to determine the rule to find the n^th term of any given sequence.

Before even attempting to determine what the n^th term of a given sequence is, we need to determine if the sequence you are given is linear, quadratic or cubic. The way to do this is by looking at the differences between the numbers in the sequence. For a linear sequence the difference between the numbers in the question will always be the same, for example

5, 9, 13, 17, …

The difference between 5 and 9 is 4. The difference between 9 and 13 is 4. And the difference between 13 and 17 is 4. Because the first difference is always the same then this is a linear sequence.

How this next sequence : -

8, 18, 32, 50, 72, …

Firstly let's look at the first differences

$18 - 8 = 10$

$32 - 18 = 14$

$50 - 32 = 18$

$72 - 50 = 22$

We can see that the first difference is gradually increasing between one number and the next. So what about the second difference (i.e. the difference between numbers in the first difference)?

$14 - 10 = 4$

$18 - 14 = 4$

$22 - 18 = 4$

In this example the second differences are the same. This is a quadratic sequence. This means that the expression for the n^th term of this sequence will contain an n² term.

If all that sounds a little confusing then here is the same thing written out in a table, which should make it a little easier to understand…

The first line in the table is the sequence itself. The second line is the first difference. The third line is the second difference. That the second difference is a constant indicates that this is a quadratic sequence.

Here is another sequence (and this is taken from a question on the YourMathsTutor forum):

0, 11, 44, 110, 220

The first differences are as follows

$11 - 0 = 11$

$44 - 11 = 33$

$110 - 44 = 66$

$220 - 110 = 110$

The second differences are…

$33 - 11 = 22$

$66 - 33 = 33$

$110 - 66 = 44$

Now we can see that in this example the second differences are still increasing, so let's look at the third differences…

$33 - 22 = 11$

$44 - 33 = 11$

The third differences are the same. This is a cubic sequence. Again, it is much easier to see this pattern if we write out a table…

Now we have a technique for recognising the basic types of sequence (linear, quadratic or cubic), we can now look at techniques for finding the rules of these sequences.

Let's take the example above,

5, 9, 13, 17, …

As the first difference is always 4 then this is a linear sequence.

The next stage of the technique is to write out a table, as follows…

n	Value	How to get there
1	5
2	9
3	13
4	17

The “How to get there” column is our key to unlocking the problem of finding the general rule for finding the n^th term of a linear sequence.

How did we get to a value of 5 for n=1? Well, that was just the value we were given. It is the start of the sequence so for n=1 we can just write 5 into the table…

n	Value	How to get there
1	5	5
2	9
3	13
4	17

What about n=2? Well that is 5+4. That is how we got to a value of 9 so we need to put that into the table…

n	Value	How to get there
1	5	5
2	9	5+4
3	13
4	17

To get the value of n=3 we take the value of n=2 and add 4 to it, in other words…

n	Value	How to get there
1	5	5
2	9	5+4
3	13	5+4+4
4	17

Likewise for n=4…

n	Value	How to get there
1	5	5
2	9	5+4
3	13	5+4+4
4	17	5+4+4+4

Hopefully you can see that n=5 will be 5+4+4+4+4…

n	Value	How to get there
1	5	5
2	9	5+4
3	13	5+4+4
4	17	5+4+4+4
5	21	5+4+4+4+4

The “How to get there” column is showing us a pattern. We can see that the general rule will look something like…

Value of n^th term = $5 +$ some multiple of 4

What do we mean by “some multiple of 4”? Looking at the sequence again we can see that when n is 2 then we have to add on 1 lot of 4. When n is 3 we have to add on 2 lots of 4. When n is 4 we add on three lots of 4. We always add on $n-1$ lots of 4. Does that work for n=1? Well, yes because 1-1=0 so we add on no lots of 4.

The rule is therefore

Value of n^th term = $5 + 4(n - 1)$

This can be simplified to

$V = 4n + 1$

NOTE: Can you see that this has the same form as the general equation of a straight line $y=mx+c$ ? If you plotted up a graph of $n$ verses $V$ you would end up plotting a straight line with a gradient of 4 that intercepts the y axis at $y=1$

The technique of tabulating the sequence as we have worked through in this example works for any linear sequence.

This time we employ a different technique. Did you realise that the rule for a linear sequence would have the same form as the general equation for a straight line ( $y = mx + c$ )? The numbers in the sequence are like points on a straight line. Well, the points in a quadratic sequence are like the points on a quadratic curve (or parabola, to give the shape its proper name). The general form of a quadratic equation is

$y = ax^2 + bx + c$

The rule for our quadratic sequence

8, 18, 32, 50, 72, …

will have this form, i.e.

$V = an^2 + bn + c$

So what we need to find are the coefficients $a$ , $b$ and $c$ .

Finding $a$ is easy - you simply take the value of the second difference (in this case 4) and divide it by 2 - $a=2$ . Note that dividing by 2 only works for quadratics 1).

Once we have found the value of $a$ we can then find the value of the n² term. Here it is tabulated…

n	V	value of $2n^2$	value of $bn + c$
1	8	2	6 (i.e. 8 - 2)
2	18	8	10 (i.e. 18 - 8)
3	32	18	14
4	50	32	18
5	72	50	22

Can you see that $bn + c$ is a linear sequence? We can find the equation for the sequence 6, 10, 14, 18, 22, … using the technique for finding linear sequences outlined above.

n	Value	How to get there
1	6	6
2	10	6+4
3	14	6+4+4
4	18	6+4+4+4
5	22	6+4+4+4+4

The equation for this linear sequence is $6+4(n-1)$ . We need to add this to the $2n^2$ term to give…

$V = 2n^2 + 6 + 4(n - 1)$

Which can be simplified to

$V = 2n^2 + 4n + 2$

The method is much the same for a cubic sequence as it is for a quadratic sequence. It is discussed in a good deal of detail on the YourMathsTutor forumhere)).

Sequences

Introduction

What sort of sequence?

Finding the rule for a linear sequence

Finding the rule for a quadratic sequence

Finding the rule for a cubic sequence