Where did this formula come from?

  • Thread starter twoflower
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  • #1
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Hi,

I wonder how could I get to this formula:

[tex]
1^2 + 2^2 + 3^2 + ... + n^2 = \frac{n(n+1)(2n+1)}{6}
[/tex]

I'd like to know it, because I suspect I will keep this formula in head and because it is quite useful I'd like to be able to write it from scratch.

Thank you.
 

Answers and Replies

  • #2
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With out working out a fair few calculations analysing the results I don't see any way to work out this formula from scratch (there probably is an easy way without working out examples, maybe I've even seen it, but I don't remeber any :frown:).

But proving it is easier. You need to use induction on n. (Do you know how to use mathematical induction as a form of proof?)
 
  • #3
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Yes, proving it with induction would be quite easy, but I can't imagine how could I think up such a formula...
 
  • #4
arildno
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twoflower said:
Yes, proving it with induction would be quite easy, but I can't imagine how could I think up such a formula...
I believe I saw an ugly formula once for an arbitrary (integral?) power "p", i.e:
[tex]\sum_{i=1}^{n}i^{p}=ugly(n,p)[/tex]
Wrap your head about that one..:wink:
 
  • #5
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Well, you say you saw somewhere how the formula came into the world? I don't even know where could I try to search it...
 
  • #6
matt grime
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Let's use it as a model of how research happens:

we know the sum of the first n numbers, it's also not hard to show the the sum of the first n cubes is acutally (the sum of the first n numbers) squared, ie a degree 4 polynomial.

conjecture:

the sum of the first n r'th powers is a a polynomial in n of degree r+1.

proof:

consider the sum of the first r+2 r'th powers show they sadisfy a general poly of degree r+1 and then prove it inductively for all n.

there is actually a more general way of demonstrating this, though i don't recall it off the top of my head
 
  • #7
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To get the formula 1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1) / 6 begin with the formula

(v+1) ^3 - v^3 = 3v^2 + 3v +1. Write down the values for v = 0, 1 , 2, ..., n and add. We obtain

(n+1)^3 = 3S2 + 3S1 + n + 1, where S1 = 1/2n(n+1).

Substituting the formula for S1, we get

3S2 = (n+1)[(n+1)^2 - 1 - 3/2n] = (n+1)(n^2 + 1/2n)

which means that S2 = 1/6n(n+1)(2n+1)


In general, if you want to get the sum of the first n cubes, fourths, fifths, etc.. by scratch, represent your sum as

(v+1) ^ n+1 - (v+1)^n for the sum of the first n whatever. Then you will have to use previous known sums (I used S1 = 1/2n(n+1)) and its easy from there. Also try proving that the sum of the first n cubes equals (1+2+...+n) ^2. Ask if you need any help
Try this technique for the sum of the first n cubes.
 
Last edited:
  • #8
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matt grime said:
...it's also not hard to show the the sum of the first n cubes is acutally (the sum of the first n numbers) squared, ie a degree 4 polynomial.

I don't see the point...

You're saying that you can express this

[tex]
1^3 + 2^3 + 3^3 + ... + n^3
[/tex]

somehow with

[tex]
1 + 2 + 3 + ... + n
[/tex]

???
 
  • #10
Galileo
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Try evaluating
[tex]\sum_n (n+1)^3-n^3[/tex]
in two ways.
 
  • #11
mathwonk
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matt grimes point was that if you think it is a polynomial of degree 3, then just calculating the first 4 values will tell you what polynomial it is. then induction will verify your guess.

courtigrads argument (for all powers) is a footnote on page 27 of the famous book Differential and integral calculus by richard courant.
 

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