Talk:Sum of Sequence of Cubes

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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. --69.116.75.131 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.--69.116.75.131 00:06, 3 January 2011 (UTC)

I made the proof more concise by just writing everything in summation form. Hope that makes things clearer. --69.116.75.131 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)