Where did this formula come from?

1. Nov 23, 2004

twoflower

Hi,

I wonder how could I get to this formula:

$$1^2 + 2^2 + 3^2 + ... + n^2 = \frac{n(n+1)(2n+1)}{6}$$

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.

2. Nov 23, 2004

Dogtanian

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 ).

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. Nov 23, 2004

twoflower

Yes, proving it with induction would be quite easy, but I can't imagine how could I think up such a formula...

4. Nov 23, 2004

arildno

I believe I saw an ugly formula once for an arbitrary (integral?) power "p", i.e:
$$\sum_{i=1}^{n}i^{p}=ugly(n,p)$$

5. Nov 23, 2004

twoflower

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. Nov 23, 2004

matt grime

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. Nov 23, 2004

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: Nov 23, 2004
8. Nov 23, 2004

twoflower

I don't see the point...

You're saying that you can express this

$$1^3 + 2^3 + 3^3 + ... + n^3$$

somehow with

$$1 + 2 + 3 + ... + n$$

???

9. Nov 23, 2004

Last edited: Nov 23, 2004
10. Nov 23, 2004

Galileo

Try evaluating
$$\sum_n (n+1)^3-n^3$$
in two ways.

11. Nov 23, 2004

mathwonk

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.

12. Nov 24, 2004