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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

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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**kylekatarn****Member**- Registered: 2005-07-24
- Posts: 445

Yes, it does.

You can try proving it by induction.

*Last edited by kylekatarn (2005-11-21 06:15:51)*

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**ryos****Member**- Registered: 2005-08-04
- Posts: 394

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

El que pega primero pega dos veces.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

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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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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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**kylekatarn****Member**- Registered: 2005-07-24
- Posts: 445

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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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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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**_________****Guest**

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.

**Dragonshade****Member**- Registered: 2008-01-16
- Posts: 147

As I said before ,

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**lashko****Member**- Registered: 2009-08-13
- Posts: 8

James wrote:

This is the text i want to quote.

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**lashko****Member**- Registered: 2009-08-13
- Posts: 8

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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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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 my horse****Member**- Registered: 2009-08-13
- Posts: 1

I love math

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**quittyqat****Member**- Registered: 2009-04-08
- Posts: 1,213

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 month. XP

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**ahmad01****Member**- Registered: 2009-09-29
- Posts: 4

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

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