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

  • Thread starter GeoMike
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  • #1
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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-
 
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Answers and Replies

  • #2
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I think it should be [tex] (k+1)^{3}-k^{3} = 3k^{2} + 3k +1 [/tex]. And then take [tex] k = 1,2,3, ..., (n-1), n [/tex]
 
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  • #3
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I think it should be [tex] (k+1)^{3}-k^{3} = 3k^{2} + 3k +1 [/tex]. And then take [tex] k = 1,2,3, ..., (n-1), n [/tex]
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-
 
  • #4
dextercioby
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I dunno, it's simply mathematical brilliancy. Basically it's intuition plus lots of work on this kind of problems: numerical formulae.

Daniel.
 
  • #5
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I dunno, it's simply mathematical brilliancy. Basically it's intuition plus lots of work on this kind of problems: numerical formulae.

Daniel.

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

-GeoMike-
 
  • #6
the equation stated to solve the equation (k+1)^3=k^3+3k^2+3k+1 comes from the algebraic proof of 1^2+2^2+3^2+4^2=(1+2+3+4)^3. This also means that 1^2+2^2+3^2+4^2.....+n^2=(1+2+3+4......+n)^3. In order to find the summation formula for the square, you need to take the cube.
 

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