Studying Mathematics Thread

[robvalue]

Yup, that is the same method I used.

To get to the next one, you do the same thing but starting with sum[(k+1)^4 - k^4]

Using the binomial theorem, that gives you (inside the sum)

k^4 + 4k^3 + 6k^2 + 4k + 1 - k^4

The k^4 cancels as before, and when you sum everything you can rearrange to find sum(k^3) in terms of sum(k^2) and sum(k) which we already know.

The proof that polymath requested that the answer is 1/p for all powers p is, I expect, the fact that pC1=p (the second coefficient in the binomial expansion is just the power p), which you end up dividing the LHS by. I'd have to write that out better to formalize it All lower powers ends up disappearing once you divide by k^p and take the limit.

Nice way to do it!

So SUM(k^p) is a polynomial in n with leading coefficient 1/(p+1).

Question: What is the next-to-leading coefficient?

After the conclusion of much scribbling, and considering the next series sum down:

[(p+1) - {(p+1)p/2}*{1/p}]/[p+1]

=[(p+1)/2] / [p+1]

=1/2