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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-
(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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