Connection between cubed binomial and summation formula proof (for squares)

I was reading through a proof of the summation formula for a sequence of consecutive squares (1² 2² + 3² + ... + n²), and the beginning of the proof states that we should take the formula:
(k+1)³ = k³ + 3k² + 3k + 1
And take "k = 1,2,3,...,n-1, n" to get n formulas which can then be manipulated into the form n(n+1)(2n+1)/6

I can follow the proof without issue, what I'm a little confused about is where "(k+1)³ = k³ + 3k² + 3k + 1" comes from. It's just given at the beginning of the proof with no logical explaination as to where it came from. I understand that using it allows us to easily get to the final form -- but is there some logical connection between the cubed binomial (and it's expansion) and the sum of the sequence of squares?

Hopefully what I'm asking makes sense...
-GeoMike-

 

Yes, they subtract k³ from each side in the next step, then take k = 1,2,3,...(n-1),n.

But, what is the process of reasoning that led us to use that equation as a starting point for the proof? For example, if the question had just said "derive a formula for the sum of the sequence of consecutive squares" or "Prove that n(n+1)(2n+1)/6 is a formula for such a sequence" I wouldn't instantly think "Oh yeah, that's easy, I just need to use this cubic equation" -- what process of reasoning would get me there?

-GeoMike-

 

I can live with that. I just wanted to make sure I wasn't missing something really obvious... :tongue2:

-GeoMike-