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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 (2005-11-21 04:57:58)
A logarithm is just a misspelled algorithm.
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I didn't learn that one till college algebra...
El que pega primero pega dos veces.
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Induction? I'd like a proof. It appears my mathbook assumes I know it. Must have been an updated version of a previous book I didn't know about.
Last edited by mikau (2005-11-21 13:07:21)
A logarithm is just a misspelled algorithm.
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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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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.
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.
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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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.
As I said before ,
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This is the text i want to quote.
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24[(n-1)²] + 2 = (2n-1)³ - (2n-3)³
24[(n)²] + 2 = (2n+1)³ - (2n-1)³
how did u get those equalities?
Thanks
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Lashko, just use the fact that:
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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I love math
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Hi, i love my horse!
P.S. Click the smiley and it will take you to a screen where it will wave.
Last edited by quittyqat (2009-08-14 02:22:06)
I'll be here at least once every decade.
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How if i^3 = [1/2 n (n+1)]^2 ???
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