1. ## Deducing polynomials technique

Hi guys,

I was hoping someone would be able to tell me the name of this technique (described below) I used so I can go away and do some reading about it and hopefully learn it in a more structured way. Also if there are any similar techniques for deducing polynomials from series (like expanding the series) could you tell me the names of them so I can do some reading about them too. Or any other techniques for finding the sum of squares. Thanks.

The technique:

So I was trying to deduce the sum of squares:

$\sum_{x=1}^{n}x^2=x^3/3+x^2/2+x/6$

and I used the technique where you go down the levels of the difference. i.e
the first few terms of this series is:
1,5,14,30,55,91,140
difference of these terms is:
4,9,16,25,36,49
and the diff. of the diff.:
5,7,9,11,13
and the level 3 diff.:
2,2,2,2
so now we can see that since it's level 3 .
$\sum_{x=1}^{n}x^2=O(x^3)$
comparing this to $x^3$ series:
series:
1,8,27,64,125,216,343
level 1:
7,19,37,61,91,127
level 2:
12,18,24,30,36
level 3:
6,6,6,6
comparing level 3's of $x^3$ and $\sum^n_{x=1}x^3$ we get:
$\sum_{x=1}^{n}x^2=x^3/3+O(x^2)$

Now the level 2 difference of $\sum_{x=1}^{n}x^2-x^3/3$ is:
1,1,1,1,1
and level 2 difference for $x^2$ is:
2,2,2,2,2
so we now have:
$\sum_{x=1}^{n}x^2=x^3/3+x^2/2+O(x)$

Finally the level 1 difference of $\sum_{x=1}^{n}x^2-x^3/3-x^2/2$ is:
1/6,1/6,1/6,1/6,1/6,1/6
and for $x$ is:
1,1,1,1,1,1

$\sum_{x=1}^{n}x^2=x^3/3+x^2/2+x/6$

and I got the right result. Now what's this technique called?

2. should be n^3/3 + n^2/2 + n/6

3. Newton's finite difference method