Find the nth term of the sequence: 14, 37, 72, 119, 178

Check the forward differences: if is the -th term, then the first forward difference of is .





The usefulness is this: if the sequence were arithmetic, then the forward differences would be constant. For example, if the sequence were , we would see the differences to be , and so the sequence would be growing linearly, or by an added constant.

In our case, we can compute the forward differences again (the second-order differences) until we find such a pattern. This time, we denote it by .



So now we know that the sequence is arithmetic with a common difference of 12 between its terms.



Now consider ; we can write this term of the sequence in terms of the previous ones. We have






We would see a similar recursive pattern if we looked at the other terms and . We then establish that the sequence is given by the recursive formula



There are lots of ways to find the explicit formula for from this point, but the simplest is to realize that the sequence must be quadratic (this is because the second-order differences are constant) so we can assume that



for some constants . We have to solve for three unknowns, so we need three known values:







So the -th term is determined by



for all .