Studying Mathematics Thread

[GrandizerII]

To prove 1+2+3+...+n=n(n+1)/2

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

In this case, S is also:
n+n-1+n-2+...+1

If you add both equations of S, you get:

2S=n+1+(n-1+2)+(n-2+3) + ... + n+1

2S = n+1 + n+1 + n+1 + ... + n+1   (n times)

Therefore S = n(n+1)/2

Therefore 1+2+3+...+n = n(n+1)/2

Now for the other two theorems stated in my previous post, how to derive them remains a puzzle to me at this stage. I can prove them via mathematical induction only.

I can show you how if you want

Yes please. Google hasn't been kind to me in this case.

---

Actually, I figured out how to derive the second theorem thanks to a post on a physics forum (though I did have to continue the derivation on my own since that poster didn't complete the derivation to the end).

Basically, it's like this:

Start with SUM(k+1)^3-k^3 for all natural numbers k up to n, and equalize it to SUM(3k^2+3k+1) for all natural numbers k up to n. From now on, when you see SUM in this post, just know it involves all instances of k (a natural number) up to n.

On the left side of the equation, you end up with (n+1)^3-1^3 (as a result of subtracting out all the other terms). On the right side, 3*SUM(k^2) + 3*SUM(k) + 1(n times).

So ...

(n+1)^3-1 = 3*SUM(k^2) + 3*SUM(k) + n
n^3+3n^2+3n+1-1-n-3*SUM(k) = 3*SUM(k^2)
(n^3+3n^2+2n)/3-SUM(k) = SUM(k^2)

At this point, we've managed to isolate SUM(k^2), and so we now focus only on the left side of the equation.

We know that SUM(k) = n(n+1)/2

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

Hence,
SUM(k^2) = n(2n+1)(n+1)/6

Now for the third theorem, I'll just leave it up to you rob, lol.

[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.

[GrandizerII]

While you do that proof requested by polymath, here's my derivation for that third theorem "on paper" (yeah, I figured it'd be something similar to the second one).

https://pasteboard.co/HfkDYKZ.png

Fuck yeah, I love maths.

[polymath257]

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?

[GrandizerII]

This one I will only very briefly ponder on for now. Improper integrals that happen to be convergent ... Wow ...

EDIT: Ok, nevermind, I can intuit it quite well now.

[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

[Silver]

Opening this thread makes my head hurt.

[robvalue]

That's kinda cool. It's always 1/2. Unless I done buggered it up.

[polymath257]

That's kinda cool. It's always 1/2. Unless I done buggered it up.

Nope. You got it right! it *is* cool, isn't it?

[robvalue]

So cool I might have to organize a party right now!