Sum of consecutive cubes

ray giraffe

The recent thread on consecutive numbers reminded me of the following fact, which is well known, but is there any elegant reason for it?

[latex]\displaystyle \left( \sum_{i=1}^n i \right)^2 = \sum_{i=1}^n i^3[/latex]

It is easy to prove by induction, but can anyone say anything better?

Edit: Thinking again, I guess it's related to

[latex]\displaystyle \left( \int_0^n \! i \, di \right)^2 = \int_0^n \! i^3 \, di[/latex], although
[latex]\displaystyle \left( \int_0^n \! 1 \, di \right)^2 = 2\left( \int_0^n \! i \, di \right)[/latex] does not imply
[latex]\displaystyle \left( \sum_{i=1}^n 1 \right)^2 = 2\left( \sum_{i=1}^n i \right)[/latex].

Fringe

This link gives a pretty cool geometric interpretation of n(n+1)/2 if that's what you're mentioning in the first part.

http://www.billthelizard.com/2009/07/six-visual-proofs_25.html

ray giraffe

Fringe wrote: »

This link gives a pretty cool geometric interpretation of n(n+1)/2 if that's what you're mentioning in the first part.

http://www.billthelizard.com/2009/07/six-visual-proofs_25.html

That's cool thank you

MathsManiac

Nice link indeed, but not quite what was being asked, I think.

Does this help?
http://en.wikipedia.org/wiki/Squared_triangular_number

The visual demonstration there takes a little too much thought to be truly elegant, I think, but it's pretty good.

And here's another:
http://users.tru.eastlink.ca/~brsears/math/oldprob.htm#s32
(Typo at the end of the first line - the expression should be squared.)

Fremen

ray giraffe wrote: »

Edit: Thinking again, I guess it's related to

[latex]\displaystyle \left( \int_0^n \! i \, di \right)^2 = \int_0^n \! i^3 \, di[/latex], although
[latex]\displaystyle \left( \int_0^n \! 1 \, di \right)^2 = 2\left( \int_0^n \! i \, di \right)[/latex] does not imply
[latex]\displaystyle \left( \sum_{i=1}^n 1 \right)^2 = 2\left( \sum_{i=1}^n i \right)[/latex].

I have a feeling that the implication works in the other direction. That is, if you assume the sums are equal, then it will follow that the integrals are equal, via

[latex]\displaystyle \int_0^n \! f(i) \, di = \sum_{k=1}^n \int_{k-1}^k f(i)di[/latex]

I tried proving it but it gets kind of messy and I should probably do some real work.

ray giraffe

MathsManiac wrote: »

Nice link indeed, but not quite what was being asked, I think.

Does this help?
http://en.wikipedia.org/wiki/Squared_triangular_number

The visual demonstration there takes a little too much thought to be truly elegant, I think, but it's pretty good.

And here's another:
http://users.tru.eastlink.ca/~brsears/math/oldprob.htm#s32
(Typo at the end of the first line - the expression should be squared.)

Both links are fascinating, thank you!

ray giraffe

Fremen wrote: »

I have a feeling that the implication works in the other direction. That is, if you assume the sums are equal, then it will follow that the integrals are equal, via

[latex]\displaystyle \int_0^n \! f(i) \, di = \sum_{k=1}^n \int_{k-1}^k f(i)di[/latex]

I tried proving it but it gets kind of messy and I should probably do some real work.

I don't see why [latex]\displaystyle \int_{k-1}^k f(i) \,di = f(k)[/latex] should be true, except that if [latex]\displaystyle f[/latex] is well-behaved then they should be asymptotically equal for large k.

Fremen

ray giraffe wrote: »

I don't see why [latex]\displaystyle \int_{k-1}^k f(i) \,di = f(k)[/latex] should be true, except that if [latex]\displaystyle f[/latex] is well-behaved then they should be asymptotically equal for large k.

I was thinking along the lines of

[latex]\displaystyle \int_0^n i di = \sum_{k=1}^n \int_{k-1}^k i di = \sum_{k=1}^n \frac{(k)^2 - (k-1)^2}{2} [/latex]

[latex]= \displaystyle \left(\sum_{k=1}^n k\right) - \frac{n}{2}[/latex]

so that
[latex]\displaystyle \left(\int_0^n i di\right)^2 = \left(\sum_{k=1}^n k\right)^2 - n \left(\sum_{k=1}^n k\right) + \frac{n^2}{4}[/latex]

now do something similar for the integral of the cube and apply the hypothesis that the sums are equal. I dunno, maybe it won't work.

Iderown

The "blood and guts" method of summing series of powers of positive integers goes like this.

Non-fully worked example:

Assume

[latex]\displaystyle \sum_{i=0}^n i^3 = a + bn + cn^2 + dn^3 + en^4[/latex]

Form the five equations using each of n = 0 to n = 4 in turn.
Note that the equation involving n = 0 immediately gives a = 0.

Starting the summation at i = 0 is exactly the same as starting at i = 1.

Solve the 4 linear equations. Simplest method is by systematic elimination, but this yields no insight about the interesting similarity between the result for n = 1 and that for n = 3.

Maybe solution by QD factorisation of the matrix of coefficients would be interesting. Likewise the eigenvalues of the coefficient matrices may be interesting. I'll save those calculations for a longer train journey.

ray giraffe

Finally found a beautiful visual proof

https://imgur.com/0dEQIAU

godtabh

Fringe wrote: »

This link gives a pretty cool geometric interpretation of n(n+1)/2 if that's what you're mentioning in the first part.

http://www.billthelizard.com/2009/07/six-visual-proofs_25.html

That's cool