Printable View
I figure why not, to start off the new year. I observed years ago that the sums of the cubes of the first N digits is the sum of those same N digits squared. You can include 0 if you like.
1^3 = 1^2 = 1
1^3 + 2^3 = (1 + 2)^2 = 9
1^3 + 2^3 + 3^3 = (1 + 2 + 3)^2 = 36
1^3 + 2^3 + 3^3 + 4^3 = (1 + 2 + 3 + 4)^2 = 100
and so on forever. However, I have never seen anyone do a good job proving the formula. Can Jemediah or anyone else do it as an exercise? :ehh:
First note a somewhat similar but simpler pattern in sums:
1 = 1 = 1*2/2
1 + 2 = 3 = 2*3/2
1 + 2 + 3 = 6 = 3*4/2
1 + ... + n = n*(n+1)/2
This can be proven inductively--in essence, induction is basically like recursion. You suppose you can solve a problem in a simpler case, and show that you can use that answer to solve it in a harder case. Specifically, here we suppose 1+...+n = n*(n+1)/2 and use that result to show a similar result is true of 1+...+n+(n+1), i.e. we show 1+...+n+(n+1) = (n+1)*(n+2)/2.
Here,
1+...+n+(n+1) = (1+...+n)+(n+1) = (n*(n+1)/2) + (n+1)
= (n+1)*(n/2 + 1) = (n+1)*(n/2 + 2/2) = (n+1)*(n+2)/2
For any fixed m, we can "go up a ladder" of inferences, starting at the trivial n=1 case which we can compute by hand:
1 = 1*(1+1)/2, so
1+2 = 2*(2+1)/2, so
1+2+3 = 3*(3+1)/2, so
..., so
1+...+m = m*(m+1)/2.
This formula can be proven more intuitively as follows. Add up the numbers from 1 to 100. Do it in groups:
(0+100) + (1+99) + (2+98) + ... + (49+51) + 50
= 50*100 + 50 = 5050 = 50*100 + 50 = 100*101/2. Counting the number of groups and accounting for the leftover middle number gives the formula, though a bit less rigorously, and you have to think about even and odd cases separately.
Now for your formula, we do something very similar. First recall that (a+b)^2 = a^2 + 2ab + b^2. Suppose 1^3 + ... + n^3 = (1+...+n)^2. Now
(1+...+n +(n+1))^2 = (1+...+n)^2 + 2*(1+...+n)(n+1) + (n+1)^2
= (1+...+n)^2 + 2*n*(n+1)/2*(n+1) + (n+1)^2
= (1+...+n)^2 + (n+1)^2 * (n+1) = (1+...+n)^2 + (n+1)^3
= (1^3 + ... + n^3) + (n+1)^3
= 1^3 + ... + (n+1)^3
and we're done. I've heard this result before though I don't think I've proven it before. I'm sure there are many proofs. This one is likely not the most elegant, though the 1+...+n = n*(n+1)/2 formula is quite well-known making the proof of your proposition short. You could probably make or find a pretty intuitive geometric proof involving cubes and squares.
For a geometric proof, see this page. For the Wikipedia article, see squared triangular number (two of my professors are in the references, even!). For a geometric generalization, see this paper. Sometimes the result you mention is called "Nichomachus' theorem". An algebraic generalization I wasn't aware of is Faulhaber's formula; that page contains several related identities including the very similar square pyramidal number formula and my first formula. For a rather interesting but somewhat advanced generalization dealing with Diophantine equation solutions, see this paper.
Truly an exceptional solution and proof, posted on the Internet by one of the greatest mathematicians I have ever witnessed or studied, perhaps the greatest of them all. :thumb:
Hah, thanks, but any ~Ph.D. math student should have been able to provide that proof without difficulty. Most serious undergrad pure math people should have, even (though sometimes I have high expectations...).Quote:
Originally Posted by Code Doc
Do not sell yourself short. Few people even know that this property of integers that I posted even exists. To be able to prove that it does exist is not easy. The ease in which you did this is remarkable.
Not many today even know that (N + 1)(N) /2 yields the sum of the first N integers. :ehh: