Mathematics is full of patterns, if it wasn't, there would be no mathematics, which you might think was a good idea, until you realise what a mess you would get into if you couldn't add, subtract, multiply etc.

What is the pattern in this sequence of numbers, 3, 6, 9, 12, 15, ... ?

The three dots at the end is the usual way of saying that the sequence goes on for ever. Each number in the sequence is called a term. So in this sequence, the first term is 3, the second term is 6 and so on.

It is easy to see that this sequence is the multiples of 3. This
can be written as a formula, *n*^{th} term = 3*n*

which is the usual way of saying that any term is 3 times its position in the sequence;

the 1st term = 3 x 1 = 3

the 4th term = 3 x 4 = 12

So for the sequence 7,14,21,28,35,...
*n*^{th} term = 7*n*

Once you have a formula for a sequence, you can work out any term in the sequence.

For the sequence* n*^{th} term = 3*n, *the 100th term = 3x100
= 300

For the sequence *n*^{th} term = 7*n, *the 34th term = 7x34
= 238

1. Work out the formula for the sequence, 4, 8, 12, 16, 20, ... and use your formula to work out the 100th term.

2. Work out the formula for the sequence, 9, 18, 27, 36, 45, ... and use your formula to work out the 29th term.

Now what about 7, 12, 17, 22, 27,...?

The numbers are increasing by 5 each
time, but the formula is not simply *n*^{th} term = 5*n*
because that gives,
5,10,15,20,25,...

But compare these two sequences and you
can see that the first one is always 2 more than the second one,
so the sequence 7,12,17,22,27,... has the formula *n*^{th}
term = 5*n+*2

This can be checked, eg. 3rd term = 5x3+2 = 17 which is correct

And the formula can be used to work out, for example, the 40th term. 40th term = 5x40+2 = 202

Work out the formula for nth term of each of these sequences. Check that the formula is correct by checking one or two of the terms in the sequence and then use the formula to work out the 47th term

3. 4,6,8,10,12,... 4. 8,11,14,17,20,... 5. 12,15,18,21,24,... |
6. 5,6,7,8,9,... 7. 7,13,19,25,31,... 8. 0,4,8,12,16,... |
Answers |

To develop methods for more complicated sequences it is necessary to use a Difference Table.

Using 7,12,17,22,27,... as an example.

n |
1 | 2 | 3 | 4 | 5 | ||||

nth term |
7 | 12 | 17 | 22 | 27 | ||||

1st difference | 5 | 5 | 5 | 5 | |||||

5n |
5 | 10 | 15 | 20 | 25 | ||||

nth term - 5n |
2 | 2 | 2 | 2 | 2 |

Subtracting successive terms gives the 1st difference of 5. As this is
the same each time, the formula will start with *n*^{th} term = 5*n.*

Then working out *5n *for each term and subtracting this from the *n*th
term, gives 2 each time completing the formula* n*^{th} term = 5*n+*2
as we found before. This might seem rather complicated, but it works and
is useful for more complicated sequences, such as 6, 9, 14, 21, 30,
...

n |
1 | 2 | 3 | 4 | 5 | ||||

n^{th} term |
6 | 9 | 14 | 21 | 30 | ||||

1st difference | 3 | 5 | 7 | 9 |

Here the 1st differences are not the same, so work out the 2nd differences, that is the differences between the differences.

n |
1 | 2 | 3 | 4 | 5 | ||||

n^{th} term |
6 | 9 | 14 | 21 | 30 | ||||

1st difference | 3 | 5 | 7 | 9 | |||||

2nd difference | 2 | 2 | 2 |

The 2nd difference is the same and this means that the formula for the
sequence will start with * n*th term = *n*^{2}

n |
1 | 2 | 3 | 4 | 5 | ||||

n^{th} term |
6 | 9 | 14 | 21 | 30 | ||||

1st difference | 3 | 5 | 7 | 9 | |||||

2nd difference | 2 | 2 | 2 | ||||||

n^{2} |
1 | 4 | 9 | 16 | 25 | ||||

n^{th} term - n^{2} |
5 | 5 | 5 | 5 | 5 |

And this gives the formula * n*^{th}
term = *n*^{2}+5

Work out the formula for these sequences. Check that your formula fits the numbers in the sequence.

9. 9, 12, 17, 24, 33, ... 11. 3, 9, 19, 33, 51, ...

10. 0, 3, 8, 15, 24, ... 12. 10, 19, 34, 55, 82, ...

If you compare your difference tables with the formulae for the sequences in
questions 9,10,11,12, you may notice that the 2nd difference is twice the
coefficient of the *n*^{2 }in the formula.

You may also be thinking that if the 2nd difference is the same, the formula
starts with *n*th term = *n*^{2}, then if the formula starts
with *n*th term = *n*^{3}, then the third differences
will be the same.

Check out this theory on, 13. 1, 8, 27, 64, 125, ... 14. 4, 11, 30, 67, 128

Now for a real challenge! What is the formula for the sequence 6, 20, 52, 108, 194, ... ?

n |
1 | 2 | 3 | 4 | 5 | ||||

nth term |
6 | 20 | 52 | 108 | 194 | ||||

1st difference | 14 | 32 | 56 | 86 | |||||

2nd difference | 18 | 24 | 30 | ||||||

3rd difference | 6 | 6 | |||||||

n^{3} |
1 | 8 | 27 | 64 | 125 | ||||

nth term - n^{3} |
5 | 12 | 25 | 44 | 69 | ||||

1st difference of these | 7 | 13 | 19 | 25 | |||||

2nd difference of these | 6 | 6 | 6 | ||||||

3n^{2} |
3 | 12 | 27 | 48 | 75 | ||||

nth term - (n^{3}+3n^{2}) |
2 | 0 | -2 | -4 | -6 | ||||

1st difference of these | -2 | -2 | -2 | -2 | |||||

-2n |
-2 | -4 | -6 | -8 | -10 | ||||

nth term - (n^{3}+3n^{2}-2n) |
4 | 4 | 4 | 4 | 4 |

So the formula is *n*th term = *n*^{3}+3*n*^{2}-2*n*+4

You may decide that with something as complicated as this that it is easier to use simultaneous equations.

To do this with the same sequence 6, 20, 52, 108, 194, ...

Start with a difference table to work out the differences to find that the formula is a cubic.

n |
1 | 2 | 3 | 4 | 5 | ||||

nth term |
6 | 20 | 52 | 108 | 194 | ||||

1st difference | 14 | 32 | 56 | 86 | |||||

2nd difference | 18 | 24 | 30 | ||||||

3rd difference | 6 | 6 |

Because the 3rd differences are the same, the formula is *n*th
term = *an*^{3}+*bn*^{2}+*cn*+*d *
where* a,b,c *and* d *are to be worked out. Any of* b,c *or*
d *may =0.

As there are four unknowns (*a,b,c *and* d* ) four equations are
needed and to get these the first four terms of the sequence are used.

When n=1 1^{3}*a*+1^{2}*b*+1*c*+*d*=6
Equation 1 *a*+*b*+*c*+*d*=6

When n=2 2^{3}*a*+2^{2}*b*+2*c*+*d*=20
Equation 2 8*a*+4*b*+2*c*+*d*=20

When n=3 3^{3}*a*+3^{2}*b*+3*c*+*d*=52
Equation 3 27*a*+9*b*+3*c*+*d*=52

When n=4 4^{3}*a*+4^{2}*b*+4*c*+*d*=108
Equation 4 64*a*+16*b*+4*c*+*d*=108

Subtracting equation 1 from each of the other 3 gives
Equation 5 7*a*+3*b*+*c*=14

Equation 6 26*a*+8*b*+2*c*=46

Equation 7 63*a*+15*b*+3*c*=102

Subtracting 2 times equations 5 from equation 6 gives
Equation 8 12*a*+2*b*=18

Subtracting 3 times equations 5 from equation 7 gives
Equation 9 42*a*+6*b*=60

And subtracting 3 times equation 8 from equation 9 gives
6*a *= 6

*a *= 1

This can now be substituted into equation 8 or 9 to work out *b *and
these values into equation 5,6 or 7 to find *c* and then these values into
one of the original equations to find *d*. Then you can check that
the formula for the nth term works on the original sequences, hoping that you
haven't made any arithmetical errors on the way!

So, if you are feeling brave, find the formula for each of the following sequences, either using the difference table method or the simultaneous equations method.

13. 4, 31, 88, 187, 340, ... 14. 6, 18, 42, 84, 150, ... 15. 6, 30, 90, 210, 420, 855, ...