S’_k = kS_(k-1) + constant

Actually the constant = (-1)^k * B(k), B(k) is the kth Bernoulli Number

B(0) = 1, B(1) = -1/2, B(2) = 1/6…

Even if you do not Bernoulli Numbers, you may use simple substitution to wok out the “constant)

using simple integration and substitution, you can then work out all kth power sums.

For example,

S0 = n, S1 = (n^2)/2 + cn (integration of S0)

put n = 1, S1(1) = 1 = (1^2)/2 + c, c= 1/2

hence, S1 = (n^2)/2 + n/2

2*S1 = (n^2) + n

integrate 2*S1 = (n^3)/3 + (n^2)/2

hence S2 = (n^3)/3 + (n^2)/2 + cn

put n = 1, c = 1/6

S2 = (n^3)/3 + (n^2)/2 + n/6 = n(n+1)(2n+1)/6

S3, S4… can be worked out similarly.

]]>0, 1, 8, 27, 64, 125, …

1, 7, 19, 37, 61, …

6, 12, 18, 24, …

6, 6, 6, 6, …

0, 0, 0, …

From this I read the first column to tell me that j^3 = 0 (j choose 0) + 1 (j choose 1) + 6 (j choose 2) + 6 (j choose 3). Then I observe on Pascal’s triangle that the sum of (j choose i) from j = 0 to n is simply (n+1 choose i+1). Thus S_k(n) = 0 (n+1 choose 1) + 1 (n+1 choose 2) + 6 (n+1 choose 3) + 6 (n+1 choose 4).

]]>*Thanks — I’ve put that right now.*

I didn’t know that . (To get the latex to appear one writes @latex 3^3+4^3+5^3=6^3@ except that instead of the @ signs one puts $ signs.)

I presume the obvious pattern doesn’t continue. In other words, I presume that it is *not* true that . Indeed, it can’t hold because the left-hand side is even.

One could also ask the much more general question of whether there are other examples of identities of the form

For that to hold, we need , which means that must be small enough that the powers are still increasing pretty fast. So we need to be of a similar order of magnitude to . This order of magnitude should be around , so that needs to be around . A back-of-envelope calculation then tells us that will have to be of similar magnitude to . But whether there are any more solutions I don’t know: I wouldn’t expect an infinite family of solutions.

In fact, I think there could be a heuristic argument that suggests that there are only finitely many solutions. Roughly speaking, for each there will be only a small number of s that could possibly work. If you now test each one, it will have a probability of … this is the bit I’m not quite sure about, but I’m pretty confident that the answer is small … something like of actually working. Summing over all gives a finite result.

]]>O I misspelled, I meant:

$$3^2 + 4^2 = 5^2$$

$$3^3 + 4^3 + 5^3 = 6^3$$

which is not only more related to the post than what I wrote, but also more true.

(Also: how do I make part of my replies into Latex if it is apparently not by typing ‘$$’?)

]]>$$3^2 + 4^2 = 5^2$$

$$3^3 + 3^4 + 3^5 = 3^6$$

?

]]>I am using the latest Firefox on a Mac; for me, the math does scale properly along with the text, when I make the text larger using cmd-+.

]]>That is indeed very easy conceptually, though the trick in the approach in the post leads to cancellation that means that it requires less calculation. Also, with this approach it looks like a miracle that ends up simplifying down to , rather than being a very natural consequence of the argument.

]]>and since most of the LHS sum of cancels against most of the RHS sum of , we can rearrange trivially to get a formula for the next term down, namely :

This approach generalises to give a formula for each in terms of all the previous ones:

and so on in an obvious pattern.

]]>It is significant, in that it can be used to give a slightly different method of proof. I’ll illustrate it with summing squares. This time, let’s take the discrete cube consisting of all points with every coordinate between 1 and , and let’s simply count points with maximal, points with maximal and points with maximal. I’ll call these sets and .

Then and each have size . The intersection of any two of them has size . And finally the intersection of all three of them has size . So by the inclusion-exclusion formula we get

That gives us that , which equals .

In general, if we do things this way, we get that

For example, , so .

This formula is perhaps simpler conceptually than the one in the post, but requires more calculation because you have to calculate all the earlier .

]]>Thanks — I’ll check that out.

]]>Not sure if that’s significant or obvious.

]]>*Thanks — I’ve corrected that now.*

Very nice proof for the sum of kth powers. Recommended for reading! ]]>

2) From the finite difference considerations it is clear that the formula for the sum of the k-th powers is a polinomial of degree k+1 in n. Actually we can just guess that from examples. Then we can figure out the coefficients of this polynomial by fitting the small n data, and then finish the proof by induction.

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