You did essentially the exact same thing in a different order.

(100+1)÷2 × 100

(100+1)x100 ÷ 2

I think you'd need to prove that the average is (100+1)/2 because that's not an axiom.

I'd say it's fifty-fifty.

The math is the same, you just wrote it more "casually". For me it was 0+100, 1+99, 2+98 ... 49+51 -> 100 x 50 = 5000, then add the 50 that was missed from the middle for 5050. But yeah I remember coming up with that when I was really young.

This is my first time seeing this problem. Interesting that they taught it in school.

Sorry if this is stupid but how to deal with sums to odd numbers ? Won't you have a number left over after pairing all the others?

This is why I never succeeded at math. Like why does this shit work?? How can people just take a problem and be like, nah I'm going to just throw numbers all over the place and reassemble them in all sorts of ways and get an answer somehow...

I can't just memorize arbitrary nonsense that "just is" I need to know how it works or it never sticks and all the math I've ever been taught was just "memorize this arbitrary nonsense and regurgitate a specific formula for a specific application that we've spent 0 time explaining other than telling you to memorize it. You want proofs and you can't get proofs until advanced college courses" well guess I'll just never understand mathematical manipulation then...

I feel like 50/50 school failed me and I failed at math.

I'm a spatial-visual person, so when presented with this problem as a teenager, I instead solved it spatially. If you stack squares like.

█.
██.
███.
...

To the hundredth row, you get a shape that is a half filled square that is 100x100. Except the diagonal is fully filled in, so you need to add another 50.

So the answer was 0.5x100x100 + 0.5x100. Easy to visualize, easy to solve. 5050.

There's a similar problem in sports -- I was a teaching assistant for our rural school's gym class so this one also popped up for me as a teenager. If you have 100 teams and each team needs to play each other team once... You fill in a similar grid, with the teams on both the x and y axis. The diagonal gets removed in this scenario because a team cannot play itself. So the answer is 0.5x100x100 - 0.5x100. 4950. Anyone who has ever tried to plan any sort of tournament can probably solve this intuitively, but 25 years ago I though I was the smartest gym class teaching assistant ever ;)

The algorithm gets a little weird if you're summing the numbers to an odd number, though since there will be a left over number you have to deal with . By calculating 2S it works exactly the same in either case.

picture a square that is n times n, grid size 1. cut it diagonaly, you have n²/2, but that leaves out the bunch of triangles that are 1/2 each. how many of these triangles are there? n.

so we are left with n²/2 + n/2.

which is (n² + n)/2.

which is n(n + 1)/2.

edit.: maybe using an irl example like counting steps on a staircase and their area could help ilustrate it better, idk.

I've always seen it explained as (100+0) + (99+1) + (98+2) + (97+3) + ...

So you basically have 50 pairs of numbers that all equal 100. So 5000 right? But there is one number that doesnt have a pair to make 100. Which is 50. Obviously there arent two 50s to make 100. So its added by itself. 5050.

Its the same as the summation of 1 to 10. (0+10) + (1+9) + (2+8) + (3+7) + (4+6) + 5. Its 55.

Addition?

There's a nice book about this: https://www.goodreads.com/book/show/57007645-thinking-better so you can learn.

I... haven't read it yet, but it's on my list. I've only listened to some interviews with the author about it.

Ahh because Euler did everything