Date: Tue, 16 Feb 1999 15:55:36 -0600 (CST)
Subject: Finding a rule for a sequence

Name: Lindsey

Level: All

Question:
I'm doing a maths investigation and i have a sequence which goes:-

13,16,25,32,45,56,73.

Our teacher told us we have to find a rule by looking at the differences of the terms until we find a constant. The first differences are:-

3,9,7,13,11,17.

The differences of these are:-

6,2,6,2,6

And the differences of these are:-

4,4,4,4.

This is a constant. However I don't know a way to find a rule from this for the nth term as it's the third row of differences so I can't use the rule for if it's the second row of differences that my teacher gave me.

Also, should the 2nd row of differences be:-

+6,-2,+6,-2,+6 ?

And therefore the third row would be:-

8,8,8,8.

I have found the nth term for when 'n' is odd (1,3,5 etc) which is (n^2)+7 and when 'n' is even it's (n^2)+9 but I need to find a general rule.

Please could you tell me how to work it out so that I could work out the rules of similar sequences.

I hope you actually get what I'm talking about as I'm not sure myself!

Thanks a lot.

Lindsey, 16
Hampshire, England

Hi Lindsay

I constructed the difference table from your sequence and got the following.

 13 16 25 32 45 56 73 3 9 7 13 11 17 6 -2 6 -2 6 -8 8 -8 8

The final row, or even the row before it, indicate that you can consider this sequence as composed of two sequences, 13,25,45,73 and 16,32,56. You can now use the successive difference technique to discover two rules, one for the odd numbered terms and one for the even terms.
Another way to approach this problem is to construct one more row in the "difference" table by computing sums rather than differences. Here is a similar example. Suppose that you have the sequence 3, 4, 9, 10, 15, 16 then the "difference" table is

 3 4 9 10 15 16 1 5 1 5 1 4 -4 4 -4 0 0 0

Notice that the last row in this table comes from sums of successive terms in the previous row. Repeating this procedure for the sequence a, b, c, d gives

 a b c d b-a c-b d-c c-2b+a d-2c+b d-c-b+a

In my example this last row is a row of zeros, thus d-c-b+a = 0 and hence d=c+b-a. So after the first three terms of the sequence any term is the sum of the two immediately preceeding terms minus the term before that.