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#1 2005-11-22 03:57:02
- mikau
- Super Member
proof of the sum of squares
I was reading proofs for the exact area under a curve, (as the number of rectangles increases without bound) and right out of no where they replaced a sum of n squared integers into a differant expression and simply said "since 1^2 + 2^2 + 3^2 +... + n^2 = n(n + 1)(2n + 1)/6." Uh... it does? I'm certain they never told me that before. I'd remember something that cool. I thought maybe they'd explain it later when I learn about definite integrals, but they just mentioned it as if I already knew it. Never heard it before.
Does anyone know the proof for this formula? Or do I need to know more about calculus before I can understand it?
Last edited by mikau (2005-11-22 03:57:58)
A logarithm is just a misspelled algorithm.
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#2 2005-11-22 05:02:08
- kylekatarn
- Power Member
Re: proof of the sum of squares
Yes, it does.
You can try proving it by induction.
Last edited by kylekatarn (2005-11-22 05:15:51)
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#5 2005-11-23 04:05:57
- mathsyperson
- Moderator
Re: proof of the sum of squares
I found a proof in my maths book. I tried scanning and uploading it, but that didn't seem to work.
Anyway, here it is:
Consider the identity 24r² + 2 ≡ (2r+1)³ - (2r-1)³
and take r = 1, 2, 3, ..., n.
24(1²) + 2 = 3³ - 1³
24(2²) + 2 = 5³ - 3³
24(3²) + 2 = 7³ - 5³
....................................
24[(n-1)²] + 2 = (2n-1)³ - (2n-3)³
24[(n)²] + 2 = (2n+1)³ - (2n-1)³
Adding all of these gives:
24[1²+2²+3²+...+(n-1)²+n²] + 2n = (2n+1)³ - 1³
That is: 24 [sum of all squares up to n²] + 2n = 8n³ + 12n² + 6n
24[sum of all squares up to n²] = 8n³ + 12n² + 4n
= 4n(2n² + 3n + 1)
[sum of all squares up to n²] = (n/6)(2n² + 3n + 1)
= (n/6)(2n+1)(n+1)
Why did the vector cross the road?
It wanted to be normal.
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#6 2005-11-23 08:08:54
- kylekatarn
- Power Member
Re: proof of the sum of squares
Induction? I'd like a proof.
Mathematical Induction provides proofs...
BTW, mathsy, its much more easier if you work with induction and ∑-forms.
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#7 2005-11-23 08:54:14
- mathsyperson
- Moderator
Re: proof of the sum of squares
kylekatarn wrote:
BTW, mathsy, its much more easier if you work with induction and ∑-forms.
True, but I think that my book's proof is quite elegant, with all the cancelling out on the right-hand side. Plus, I'd had a hard day at school, so I didn't want to think.
kylekatarn also wrote:
Induction? I'd like a proof.
Mathematical Induction provides proofs...
Also true. The proof by induction would go something like this:
First, prove that ((n(n+1)(2n+1))/6 = 0² when n = 0.
((0(0+1)(2*0+1))/6 = 0²
So far, so good. Now we need to prove that the difference when you increase n by 1 is identical in both cases.
The difference when increasing from k to k+1 on the right-hand side is obviously (k+1)².
The left-hand side is slightly trickier.
When the value is k, it would be ((k(k+1)(2k+1))/6 = (2k³ + 3k² + k)/6
When the value is (k+1), it would be (([k+1]([k+1]+1)(2[k+1]+1))/6 = ((k+1)(k+2)(2k+3))/6 = (2k³ + 9k² + 13k + 6)/6
The difference of these is (6k² + 12k + 6)/6, or k² + 2k + 1.
ZOMG PERFECT SQUARE LOLZ. 
k² + 2k + 1 = (k+1)², which is the same result that we wanted from above.
Finally, kylekatarn wrote:
Less words, more math.
[2b | ~2b]?
Well, he did...
Why did the vector cross the road?
It wanted to be normal.
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#8 2008-05-01 21:17:52
- _________
- Guest
Re: proof of the sum of squares
hi,could you please help me?
I am looking for the proof that any polynomial with complex coefficients can be written as a sum of squares.
#14 2009-08-15 00:19:29
- quittyqat
- Power Member
Re: proof of the sum of squares
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Last edited by quittyqat (2009-08-15 00:22:06)
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