# Neat summation

1. Mar 25, 2008

### Loren Booda

Given nonzero whole numbers n, prove

13+23+33+...+n3=(1+2+...n)2

I figured this out numerically, but lack the skills to solve it analytically (no doubt by induction) and could not find it in my table of summations. I'm too old for this to be homework.

2. Mar 25, 2008

### ice109

http://en.wikipedia.org/wiki/Faulhaber's_formula

look at the case p=3

3. Mar 25, 2008

### K.J.Healey

4. Mar 25, 2008

### Loren Booda

Thanks, ice109. I'm surprised that I've never heard of Faulhaber before. I guess Bernoulli got all the acclaim.

I still don't see a derivation of my finding, however. I thought in the case I presented that p=3 on one side of the equation, and p=2 on the other, as opposed to p=3 for both sides of Faulhaber's formula.

K.J.Healey seems to have what I seek, including a proof of Nicomachus's theorem.

5. Mar 25, 2008

### ice109

i like that one better

what the hell? every n^3 is the sum of n consecutive odd numbers? wheatstone's proof seems to imply that

Last edited: Mar 26, 2008
6. Mar 26, 2008

### Gib Z

If one knows the closed forms for all cases p< n, then the closed form for p=n can be derived as such:

Set up a table into two columns, LHS and RHS of the following equation;

$$(x+1)^{n+1} - x^{n+1} = (^{n+1}C_1)x^n + (^{n+1}C_2)x^{n-1} ...+1$$.

Sum this expression for k=1, 2,3,4....m. The LHS is a telescoping series. The RHS is the sum of cases p=0, 1, 2, 3... n. Replace every series with its known closed form. Then isolate the p=n case onto one side of the equation, and simplify.

Since we only need this for up to p=3, it shouldn't be very hard.