formula of a sequence consistent difference between differences

`a = (\text(2nd level difference))/2`   

`b = \text(1st level difference) - 3a`

`c = \text(1st term) - a - b`

Or

`a = (2D)/2`   

`b = 1D - 3a`   

 `c = \text(1st term) - a - b`

Use the following mnemonic to remember these terms:

`a = (2D)/2` The ape walked her two dogs over the 2 hills.

`b = 1D - 3a` The bee pulled her dog in one direction but had to take the three anchors as well.. 

`c = \text(1st term) - a - b` The cat was at it's 1st term at school but should have taken off her apron and taken off her bonnet. 

If you want to find the formula of a sequence such as

`5, 14, 27, 44, 63, 90`

Then first work out if there is a sequence pattern.

You can see see a consistent difference on the second level i.e. a consistent different between differences.

We want to find the formula in a form `an^2 + bn + c` where `n` is the number of the sequence position (term). Use the following method: 

`a = (\text(second level difference))/2`

`b = \text(first level difference) - 3a`

`c = \text(1st term) - a - b`

Note: When we say second level difference, we mean second level difference between 1st term and 2nd term.

Note: When we say first level difference we mean first level difference between 1st term and 2nd term.

`a = (2D)/2`   

`b = 1D - 3a`   

 `c = \text(1st term) - a - b`

So the example above would give:

`a = (\text(2nd level difference))/2 = (13 - 9)/2 = 4/2 = 2` 

`b = \text(1st level difference) - 3a`

`= (14 - 5) - 3` x `2 = 9 - 6 = 3`

`b = 3`

`c = \text(1st term) - a - b = 5 - 2 - 3 = 0`

`c = 0`

`an^2 + bn + c`

`2n^2 + 3n + 0`

Ans: the sequence formula is `2n^2 + 3n`