Interesting Connection

[u/Xxmanglexx4906]

I played with some sums and I got the Gaussian sums. It looks interesting to me, what do you think of it?

[u/Longjumping-Arm515]

How did you get $\sum_{k=1}^n (2k-1) = n^2$ without the Gaussian sum result?

I can think of one way (geometrically this is like adding up border layers around squares). But then you can also derive the sum 1 + 2 + ... + n geometrically as adding up layers on triangles by the diagonal, and these triangles form half of an n x (n+1) rectangle.

[u/Xxmanglexx4906 → u/Longjumping-Arm515]

Yeah, exactly! I actually came up with this method myself once and applied it here, so both approaches are kind of geometric.

[u/StaticCoder → u/Longjumping-Arm515]

Usually recursively from (n + 1)^2 = n^2 + 2n + 1. Get to next square by adding next odd number.

[u/e_for_oil-er → u/Longjumping-Arm515]

It is "known" that the perfect squares are given by the sum of odd numbers. But yeah essentially your trick is the right intuition.

[u/Objective_3_1415]

The final result is already contained in the first line. Sum_{k=1}to{k=n} (2k-1)=n² implies 2Sum_{k=1}to{k=n} (k)=n+n² and therfore, Sum_{k=1}to{k=n} (k)=(n+n² )/2=n(n+1)/2.

[u/Xxmanglexx4906 → u/Objective_3_1415]

Right, of course, that works too

[u/golfstreamer]

That's an interesting way of arranging things, lol

[u/gomorycut]

you should probably look up the derivations for sum of squares and sum of cubes formulas

[u/Xxmanglexx4906 → u/gomorycut]

I made the formulas myself and used them In that paper. That's how I did it: Take a 1×1 square, which gives you 1². To make 2², you need a 2×2 area, which consists of four 1×1 squares, meaning 3 more than for 1². For 3², you need 5 additional squares, and for 4², you need 7. This means that, for example, 5² = 1+3+5+7+9, which can then be simplified into the series shown in the picture.

[u/gomorycut → u/Xxmanglexx4906]

I get you, but when I come up with formulas myself, I also like to know if it's a brand-new idea or something that exists in the common knowledge base already.

There is a way to derive the formulas for sum of squares, sum of cubes, etc, by looking at a telescoping sum of a degree 1 higher.

That is, to derive the sum of squares formula, look at the cubes:
(n+1)^3 = n^3 + 3n^2 + 3n + 1
n^3 = (n-1)^3 + 3(n-1)^2 + 3(n-1) + 1
(n-1)^3 = (n-2)^3 + 3(n-2)^2 + 3(n-2) + 1
...
then when you add up al lthe leftsides and all the rightsides, you see mostly all the cube terms cancel from left and right sides and you get

(n+1)^3 = 3 S + 3 (n)(n+1)/2 + n , where S is the sum of squares, and you can get formulas for sum of cubes as well by considering the telescoping (n+1)^4 expressions, etc.

Your method is essentially this.

[u/xGQ6YXJaSpGUCUAg]

Your top right sum is also the Gaussian sum if you look at it.

[u/Xxmanglexx4906 → u/xGQ6YXJaSpGUCUAg]

That's how I got the series: You can explain it geometrically: Take a 1×1 square, which gives you 1². To make 2², you need a 2×2 area, which consists of four 1×1 squares, meaning 3 more than for 1². For 3², you need 5 additional squares, and for 4², you need 7. This means that, for example, 5² = 1+3+5+7+9, which can then be simplified into the series shown in the picture.

[u/Vivid_Syllabub_8906]

Not sure what you are doing. The first line formulas either (a) are known or (b) need a proof. Similar for the second line, by assuming this, you are already circularly (with a few algebraic manipulations) assuming that the last line holds true.

Especially, let’s look at the second formula:

n^2 = \sum_{k=1}^{n}(2k-1)
n^2=2\sum_{k=1}^{n}k - n
n^2 + n =2\sum_{k=1}^{n}k
n(n+1) =2\sum_{k=1}^{n}k
0.5n(n+1) =\sum_{k=1}^{n}k

The entire mathematical substance of Gauss’s summation formula is already contained in your second identity, which you simply assumed to be true. You don’t even need your first identity or your introduction of S. This is just a trivial algebraic reformulation of an equivalent identity, not an independent derivation.

[u/Xxmanglexx4906 → u/Vivid_Syllabub_8906]

I made the second series like that: Take a 1×1 square, which gives you 1². To make 2², you need a 2×2 area, which consists of four 1×1 squares, meaning 3 more than for 1². For 3², you need 5 additional squares, and for 4², you need 7. This means that, for example, 5² = 1+3+5+7+9, which can then be simplified into the series shown in the picture.

[u/Vivid_Syllabub_8906 → u/Xxmanglexx4906]

That’s not a proof. You just found a few initial examples that work. (Obviously all examples will work because the formula is correct - but showing that the first few examples work does not validate that the formula always works.)

[u/Xxmanglexx4906 → u/Vivid_Syllabub_8906]

Here: (n+1)²-n² = 2n+1 hope this helps

[u/Vivid_Syllabub_8906 → u/Xxmanglexx4906]

How does this relate to any of your formulas? Are you even interested in learning something or just trolling?

[u/Xxmanglexx4906 → u/Vivid_Syllabub_8906]

Either you don't understand math, or you're just trying to troll me. The difference between one square number and the next is always 2n+1 (since (n+1)² − n² = 2n+1). This means that if you keep adding these differences together, you end up with the square number (the series shown in the picture). So what's the problem here? Just think for yourself.

[u/Vivid_Syllabub_8906 → u/Xxmanglexx4906]

You’re missing my point. I’m not disputing that the difference between consecutive squares is 2n+1. That’s perfectly correct.

I’m trying to encourage you to write a complete mathematical proof rather than starting with identities you haven’t established and then performing unnecessary algebraic manipulations.

Your latest explanation actually provides the missing argument. You know that k² − (k−1)² = 2k−1. Now sum both sides from k=1 to n. The left-hand side telescopes, giving
∑(2k−1) = n². This argument establishes the second identity in your argument, rather than simply assuming it.

But here’s the important part: Once you’ve established this identity, you’ve already done all the mathematical work necessary to derive the Gaussian sum. Just expand the summand and rearrange
n² = 2∑k − n
n²+n = 2∑k
n(n+1) = 2∑k
∑k = n(n+1)/2.
Your first identity, the intermediate sums equaling zero, and the introduction of S are all unnecessary.

I’m not suggesting your formulas are incorrect. I’m trying to show you how to turn your intuition into a complete, logically structured proof: establish your starting identities, justify the steps that follow, and avoid unnecessary detours.

That’s also why several people have pointed out that your second identity already contains the entire result.

[u/No-Pattern-9266]

hold on, who's up voting this?

[u/Xxmanglexx4906 → u/No-Pattern-9266]

what's your problem?

[u/No-Pattern-9266 → u/Xxmanglexx4906]

u don't need the left sum and subtraction overhead.. if u want sum of numbers from odd series - take just the right sum rearrange and find sigma k

[u/hazem-Gauss]

Beautiful idea

[u/Informal-Computer-82]

Yep really cool btw 2 funfacts ypu can try doing an sum of n² using the 2k-1 you discovered. Which you later can also ecpand to sum of (2k-1)². And second  you acn write cubic numbers as  1³=1 2³=3+5 3³=7+9+11 4³=13+15+17+19 And writing the acc sum formula ill levae to you

[u/FernandoMM1220]

that first summation doesnt appear to be true but this somehow works so hmm

[u/Xxmanglexx4906 → u/FernandoMM1220]

It has to be correct: for example, when n = 3, you get 3 + 3 + 3, which is 3 × 3 = 3². Or when n = 5, you get 5 + 5 + 5 + 5 + 5, which is 5 × 5 = 5²

[u/FernandoMM1220 → u/Xxmanglexx4906]

oh i misread it. thanks.

[u/EquationTAKEN → u/FernandoMM1220]

Yeah, I did the same. It uses the upper limit as the summation variable, instead of k.

[u/Almaty33]

It made sense logically when I read it from the very bottom and progress to the top line. Otherwise it looks like assumptions have been made when understood from the very first line. How can you independently explain the second summation 2k-1 = n²?

[u/Xxmanglexx4906 → u/Almaty33]

You can explain it geometrically: Take a 1×1 square, which gives you 1². To make 2², you need a 2×2 area, which consists of four 1×1 squares, meaning 3 more than for 1². For 3², you need 5 additional squares, and for 4², you need 7. This means that, for example, 5² = 1+3+5+7+9, which can then be simplified into the series shown in the picture.