A sequence is made up of terms, usually separated by spaces. Terms are referenced by counting (the first term is called the first term !!) and the count is kept with a variable, such as n. Let’s start with something really simple, like
4 7 10 13 16 ….
you can see the numbers go up by three each time. I say this as “the first difference is a constant and this is three”. Each time the count [I shall use an n for this] goes up (e.g. from the third term to the fourth) by one, the sequence goes up by three, so the formula must include a 3n in it. If you work out the zeroth term, it will be 1 and this also needs including, so you have that the formula for the series is 3n + 1. Check this for the third term (i.e. when n is 3: 3x3+1 = 10) and the fourth term (n is 4: 3x4+1=13). It looks like it works.
So the technique is (for a simple pattern like this one, called a linear sequence).:
i) write the differences and check they’re constant. This difference is multiplied by n.
ii) work out the zeroth term and add that onto your formula
iii) check that it works for n=1, n=2 and n= larger (like 5).
iv) if these all check, you’re probably right.
Exercise A: You try these, of the same level of difficulty:
1. 5 9 13 17 21 ..... 6. 21 32 43 54 ...
2. 6 13 20 27 34 .... 7. 120 220 320 420 ...
3. 7 6 5 4 3 ... 8. 143 100 57 14 -29 ...
4. 38 35 32 29 26 .... 9. 1007 814 621 428 ...
5. 93 86 79 72 65 .... 10. 878 755 632 509 386 ...
All of these formulae have the form ax+b. y=ax+b is a straight line graph, which is why the sequences are called linear sequences. Obvious, isn’t it?
If we had a more difficult formula, like ax²+bx+c, then how would the differences behave?
Look at just n² 1, 4, 9, 16, 25, …. The differences are the odd numbers and the second difference is constant (and two). Write a table to check this out. I say that (from lots of practice) the second difference will be 2a, for any an²+bn+c [but I can’t prove that until you have some calculus from Sixth Form Maths]. However, all the examples you write will fit this rule, so you will see that it works.
So, here’s a harder problem to look at: 7 11 17 25 35 47 59 ....
If you write the differences, this is what happens, working up the page:
2nd Difference 2 2 2 2 2 These are constant
1st Difference 4 6 8 10 12 14 These go up in twos
Sequence 7 11 17 25 35 47 61 ….
Count 0 1 2 3 4 5 6 7 These are the values of n
You write this. Work out the zeroth term, simply by extending the pattern to the left.
You can see that the zeroth term must be 5. From what we’ve done so far, whatever the formula, when n is zero, this must be the value then. In an²+bn+c, when n=0 you get just c, which is the zeroth term.
You can see that the second difference is two, so the a in an²+bn+c is 1. So we have a formula of n²+bn+5 so far – we need to find just b. To do this we look at n=1, where n²+bn+5 = 7. So 12+bx1+5 = 1 + b + 5 = 6 + b = 7 which means b is also one, and our formula is n²+n+5. Check this (always test a formula) on n = 2 and n = 7.
Any bigger n will do for a check, but a good one to check is the next one you didn’t already work out. In this example, n=8 would be good, so use the formula to predict the value when n=8, and then use the table to check what the eighth term should be. n²+n+5 when n=8 is 8²+8+5 = 77. Look at the table: we need to add 16 to 61, which is also 77. So this formula works.
The advanced technique is:
i) write the differences until they’re constant. The row that is constant gives you the order of the function; if the fifth line of differences is where the constant is, your formula begins with n⁵. This would be pretty horrid to work out, and I have an Excel worksheet on the school intranet to do this harder type for you. Even cubics are pretty nasty to lay out, so I’ll stick to squares on this sheet.
If the second difference is constant, this is 2a for the formula an²+bn+c. If the first difference is constant, then you have b (and a=0).
ii) work out the zeroth term. This is the c in an²+bn+c
iii) look what happens when n = 1 and use this to work out b if you don’t yet know it.
iv) check your formula works for n=2 and n= larger. If these check you’re probably right.
Examples for you to try:
1. 5 9 15 23 33 ..... 6. 21 32 45 60 77 ...
2. 6 16 30 48 70 .... 7. 120 200 300 420 ...
3. 66 60 52 42 ... 8. 144 120 100 84 72 ...
4. 68 65 60 53 44 .... 9. 992 892 800 716 ...
5. 92 80 64 44 20 .... 10. -88 -68 -40 -4 40 ...
Cubics, an³+ bn²+cn+d, have a third difference of 6a. You need to try n=1 and n=2 to find b and c; d is the new zeroth term. One way to make this a little easier is to find a, and then take an³ away from your problem sequence, leaving you with bn²+cn+d, which you now know how to solve. This method is what I would recommend for harder sequence: find the highest order component of the problem and subtract it from the current problem. This makes the solution a process with built-in repetition, which builds confidence. It also let’s you find mistakes in your arithmetic — if you’ve just worked out the 4th power component of your problem, there shouldn’t be a (non-zero) 4th difference in the next bit of the problem!!
Extension work:
What is the constant difference for quartics and quintics? Look at the sequence formed by the multipliers for each new a in anr as the problems get more difficult. What is the rule? What is special about this sequence of multipliers that makes it fundamentally different from those that this sheet is about?
Being able to predict numbers in a sequence (by which I mean numbers following some pattern) is a fundamentally useful skill used surprisingly often outside what you might think of as Mathematics. Being able to predict a trend is a saleable skill. Big money.
