Interesting, my teacher taught the class sequences and how to solve them, such as tables with the terms (n) and the value (f(n))

She showed us something I though was strange:

Code:

__
n | 1 | 2 | 3 | 4 | 5 __
f(n)| 3 | 17 | 55 | 129 | 251

Then she said find the difference in the values:

Code:

__
n | 1 | 2 | 3 | 4 | 5 __
f(n)| 3 | 17 | 55 | 129 | 251
14 38 74 122

Do it again, until you have a common difference:

Code:

__
n | 1 | 2 | 3 | 4 | 5 __
f(n)| 3 | 17 | 55 | 129 | 251
14 38 74 122
24 36 48
12 12

Then count the number of times you took the difference:

Code:

__
n | 1 | 2 | 3 | 4 | 5 __
f(n)| 3 | 17 | 55 | 129 | 251
14 38 74 122 :
24 36 48 : 3
12 12 :

Then she said that this means the sequence is a cubed equation (because it is three rows down).

I looked over it and it works (I think) for all exponents.

I then discovered that if the equation was in the form: $\displaystyle f(n)=an^x+b$

That the last term (the common difference) would be: $\displaystyle x!\times a$

How would someone prove this?

BTW, the equation for the tables was: $\displaystyle f(n)=2n^3+1$