The initial introduction to this method should be through the analysis of sequences whose 1st differences are constant.Confidence and understanding gained with these more familiar sequences should help the student to understand more complicated situations. The essence of the method is a comparison beween the numerical difference table and the algebraic difference table. Let's illustrate with a simpler example.
1st differences constant :
Consider the sequence 5, 11, 17, 23 ... There is a constant difference between the terms of this sequence and therefore the n th term formula will be of the form an +b a linear expression.
Comparison of the numerical difference table with the algebraic difference table allows values to be assigned: a = 6 and since a +b = 5 then b = -1. The n th term formula is 6n -1.
Now let's tackle a harder situation
1st differences constant :
Consider the sequence 15, 35, 63, 99 ... In this case there is not a constant difference between successive terms. However a difference table reveals that there is a constant difference between the differences between successive terms. In this case you say that the 2nd differences are constant. The n th term formula is of the form an 2+bn +c , a quadratic expression. Here are the numerical and algebraic difference tables for comparison:
You can now work up the first entries in the rows from the bottom row to the top row, equating and substituing to calculate the values of a , b and c . This gives a = 4 , b = 8 and c = 3 and so the n th term formula is 4n 2 + 8n + 3 .
Should the 3rd differences be constant then the resulting n th term formula will be cubic. Try the technique out on the following sequence: 3, 4, 6, 12, 25, 48 ...