The sum of the consecutive integers from 1 to n is (n/2)(n+1), as discovered (I think) by Gauss a long time ago.

The sum of the cubes of consecutive integers from 1 to n is the square of the result above.

My question is why does this result happen??

To clarify - I am 100% the result is true - proving either result by induction is a standard question for my A Level Mathematicians, and the Gauss proof is a great introduction to proof for younger students.

What I'd like to find is a

The sum of the cubes of consecutive integers from 1 to n is the square of the result above.

My question is why does this result happen??

To clarify - I am 100% the result is true - proving either result by induction is a standard question for my A Level Mathematicians, and the Gauss proof is a great introduction to proof for younger students.

What I'd like to find is a

**geometric**argument for why they are the same, rather than an algebraic one.View 12 previous comments

- Yeah, since the sums were finite, Uffe proofed it properly by induction.Jan 20, 2013
- Here is a link to a proof-by-picture on my blog: http://blog.felixbreuer.net/2013/01/20/proof-by-picture.htmlJan 21, 2013
- nice!!!Jan 21, 2013
- good yaarJan 21, 2013
- I think perhaps this is the nicest "proof by picture" of the identity. Certainly it doesn't require any 3D visualisation.

http://nrich.maths.org/325/solutionJan 21, 2013 - Here is an algebraic proof, if you are not interested then don't read at all.

Let S=1^3+2^3+3^2+...+n^3

We know that x^4-(x-1)^4=4x^3-6x^2+4x-1

Putting x=n,n-1,n-2,...,3,2,1 successively in this identity, we get

n^4 - (n-1)^4 = 4n^3 - 6n^2 + 4n - 1

(n-1)^4 - (n-2)^4 = 4(n-1)^3 - 6(n-1)^2 + 4(n-1) - 1

(n-2)^4 - (n-3)^4 = 4(n-2)^3 - 6(n-2)^2 + 4(n-2) - 1

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.

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3^4 - 2^4 = 4(3)^3 - 6(3)^2 + 4(3)^2 - 1

2^4 - 1^4 = 4(2)^3 - 6(2)^2 + 4(2)^2 - 1

1^4 - 0^4 = 4(1)^3 - 6(1)^2 + 4(1)^2 - 1

On adding them we get,

n^4 = 4(1^3+2^3+..+n^3) - 6(1^2+2^2+..+n^2) + 4(1+2+..+n) + n

=>n^4 = 4S - 6*n(n+1)(2n+1)/6 + 4*n(n+1)/2 + n

On solving,

S = (n(n+1)/2)^2

[ Note:1^2+2^2+..+n^2=n(n+1)(2n+1)/6 ]Jan 26, 2013