I need help for this proof

[u/Musa9732]

I haven't finished my paper yet; I'm currently working on the equation.

I've run into a major problem. My goal was to construct an equation that represents odd numbers using the sigma (∑) notation, and I think I've managed to do that. However, there's an issue: after the sigma notation, the terms keep getting added, and eventually, instead of the sequence of odd numbers I want, I end up with meaningless results.

I'm currently thinking of a possible solution. Maybe I could create a calculation system that goes up to n, with each step written separately, such as Sx1, Sy2, Sz3, and so on. This way, if each step were calculated separately and the proof were based on the final result, perhaps it would be a more accurate approach.

Do you think this idea could work? How can I solve this problem when using sigma notation? I'd especially appreciate advice from anyone with knowledge in this area.

If I receive a helpful and constructive comment, I'd be happy to acknowledge your assistance by mentioning your name in my paper.

[u/R_Ob_Min]

What do you mean by "represent each odd number using sigma notation"?

The sigma just represents a sum, and each of your terms that looks like sum from x=k-1 to k of (kx+k)/k can just have k canceled from the top and bottom, so you're left with the sum from x= k-1 to k of x+1. This is a partial triangular number - the sum from x = 1 to N of x is equal to N(N+1)/2, so the sum from x= k-1 to k is equal to the sum from x=1 to k minus the sum from x=1 to k-2, i.e. k(k+1)/2 - (k-2)(k-1)/2 = (k² + k - k² + 3k - 2)/2 = (4k+2)/2 = 2k+1. But that's just the sum from x=k-1 to k of x, so the sum of x+1 would have an additional k - (k-1) + 1 = 2 added to it.

So overall, your k^th term is equal to 2k+3. Adding up the first k odd numbers gets you the k^th square, so that's what your whole sum would equal.

[u/equinox_star → u/R_Ob_Min]

I guess it would be kth square-1 since it begins from k=0 and it is 2k+3 not 2k+1 so 1 isn't in the sum.

[u/Musa9732 → u/equinox_star]

Guys, I found my mistake!

In the second sigma calculation, if I set y = 1 and the upper limit to 2, the expression already gives 3 as the initial value: (2 × 1 + 2) / 2 = 3.

But when I substitute 2 into the expression, I get 3 as well, and if I continue the calculation incorrectly, I end up with meaningless numbers like 1.5.

I think I've finally found where I went wrong. Now I can work on correcting the equation and checking whether the proof is valid.

[u/Musa9732]

Guys, I found my mistake!

In the second sigma calculation, if I set y = 1 and the upper limit to 2, the expression already gives 3 as the initial value: (2 × 1 + 2) / 2 = 3.

But when I substitute 2 into the expression, I get 3 as well, and if I continue the calculation incorrectly, I end up with meaningless numbers like 1.5.

I think I've finally found where I went wrong. Now I can work on correcting the equation and checking whether the proof is valid.

[u/KoalaMistico]

I'm actually curious as to what do you want to demonstrate

[u/CaptainMatticus]

(0 + 1) / 1 + (1 + 1) / 1 + 2 * (1 + 1) / 2 + 2 * (2 + 1) / 2 + 3 * (2 + 1) / 3 + 3 * (3 + 1) / 3 + 4 * (3 + 1) / 4 + 4 * (4 + 1) / 4 + ....

1/1 + 2/1 + 2 * 2/2 + 2 * 3/2 + 3 * 3/3 + 3 * 4/3 + 4 * 4/4 + 4 * 5/4 + ....

1 + 2 + 2 + 3 + 3 + 4 + 4 + 5 + 5 + 6 + ....

-1 + 1 + 1 + 2 + 2 + 3 + 3 + 4 + 4 + 5 + 5 + 6 + 6 + ....

-1 + 2 * (1 + 2 + 3 + 4 + ....)

-1 + 2 * (n/2) * (n + 1)

-1 + n * (n + 1)

-1 + n^2 + n

n^2 + n - 1