Talk:Sum of Sequence of Cubes

From ProofWiki
Jump to navigation Jump to search

If someone wants to try formating polynomial division for this, be my guest. Otherwise I'll get back to it when I have time. --Cynic 23:24, 27 April 2008 (UTC)

Your proof is already complete. It's really nothing to claim the factorization of a 4th-degree polynomial. If people want to verify it, they can expand the $(n+1)^2(n+2)^2$ themselves. -Grambottle

Use this

--~~~~ Joe 00:19, 28 April 2008 (UTC)

I just put together the direct proof by recursion. I initially did the same concept for triangular numbers several months ago, but figured that I could do it for any sum of sequence of odd powers, so cubes was the easiest after triangular numbers. Yeah I know, it's a pretty terse and hard to look at, but whatever. -- 22:58, 2 January 2011 (UTC)

In response to the criticism of the proof at the bottom of the page, I'm just regrouping the terms and that's where the $n^4$ comes from. I'm adding $n^3$, $n$ times. Also, isn't the definition of the sum of sequence of cubes $S(n) = n^3 + S(n-1)$? I don't see why that needs to be justified.-- 00:06, 3 January 2011 (UTC)

I made the proof more concise by just writing everything in summation form. Hope that makes things clearer. -- 00:14, 3 January 2011 (UTC)

So much better. It's possible to see what's going on now. Good job. --prime mover 06:17, 3 January 2011 (UTC)