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#1 2005-11-22 03:57:02

mikau
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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.

#2 2005-11-22 05:02:08

kylekatarn
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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)

#3 2005-11-22 09:43:28

ryos
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Re: proof of the sum of squares

I didn't learn that one till college algebra...


El que pega primero pega dos veces.

#4 2005-11-22 12:07:04

mikau
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Re: proof of the sum of squares

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 (2005-11-22 12:07:21)


A logarithm is just a misspelled algorithm.

#5 2005-11-23 04:05:57

mathsyperson
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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.

#6 2005-11-23 08:08:54

kylekatarn
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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.

#7 2005-11-23 08:54:14

mathsyperson
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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. big_smile lol big_smile tongue

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.

#8 2008-05-01 21:17:52

_________
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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.

#9 2008-05-02 01:19:25

Dragonshade
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Re: proof of the sum of squares

As I said before ,

#10 2009-08-14 08:26:56

lashko
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Re: proof of the sum of squares

James wrote:

This is the text i want to quote.

#11 2009-08-14 08:29:02

lashko
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Re: proof of the sum of squares

24[(n-1)] + 2 = (2n-1) - (2n-3)
24[(n)] + 2 = (2n+1) - (2n-1)

how did u get those equalities?
Thanks

#12 2009-08-14 08:46:31

Ricky
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Re: proof of the sum of squares

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..."

#13 2009-08-14 12:31:47

i love my horse
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Re: proof of the sum of squares

I love math

#14 2009-08-15 00:19:29

quittyqat
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Re: proof of the sum of squares

Hi, i love my horse!
P.S. Click the smiley and it will take you to a screen where it will wave.


Uploaded Images
View Image: hi.gif      

Last edited by quittyqat (2009-08-15 00:22:06)


I'll be here at least once every month. XP

#15 2009-09-30 10:36:58

ahmad01
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Re: proof of the sum of squares

How if i^3 = [1/2 n (n+1)]^2 ??? roflol

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