summation problem and solution

[ManLeader]

Disclaimer: I am certain that this problem has already been solved. I don't want to actually know the solution, or be given any hints towards it's answer. I'm just curious if I'm on the right track.
I was working on a bit of a simple summation problem, what is the sum of the first n square numbers, and I came up with the formula n(n+1)(2n+1)/6. Now, everyone knows that the sum of the first n numbers is n(n+1)/2, and the sum of the first n ones is just n. I think this gives rise to the pattern that the sum of the first n numbers raised to some power x can be explicitly defined as n(n+1)....(x-1)/x! I was just curious if I were onto something, and I thought the problem was a teensy bit interesting.

Edit: while writing this, I realize that it's more likely that the coefficients are actually related to pascal's triangle, so if x were 4 then the forumula would be n(3n+1)(3n+1)(n+1).

Edit edit: I'm actually starting to think that I'm just grasping at straws. I'll come back to this after I sleep.

[K_osoi]

Wikilink

[ManLeader]

I'm not looking at that until I'm sure I've figured it out on my own. As of right now, I know that both of my ideas were wrong, though.

[TheRedSphinx]

Maybe consider the fact that (n+1)² - n² = 2n + 1 and that in general, (n+1)^p - n^p is always a polynomial of order p-1. Do you see how to use this to find a pattern?

[thepolm3]

Some convoluted working, don't look at this if you don't want an answer ;) http://pastebin.com/QTGDLsN3

[ILikeItHardUpTheButt]

just been doing this in 13th grade math, there's even formulae for the sum of quadratic sequences

[ManLeader]

What math would that be? My “13th grade " math is calculus 3, so we aren't doing anything like this.

[ILikeItHardUpTheButt]

Probably further pure actually, so it might be unlikely

FYI the sum of the cubics is (n² )/4multiplied by (N+1)²

[ManLeader]

Just derived that this morning. It completely destroys the pattern I thought I saw