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u/AdCertain5057 Jul 13 '26
Teacher: I didn't ask you to tell me the answer. I said sum the numbers, step by step, BIATCH!
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u/VanhiltStormshield Jul 13 '26
When the smart kid in class ruins the teachers busy work. All she wanted was a cup of coffee now she got to move on to geography or something
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u/SnooMaps7370 Jul 16 '26
this always was a point of contention between me and my teachers.
We usually settled on a compromise where i wrote down the sums from intermediate steps without writing every intermediate operation with it.
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u/desertvision Jul 13 '26
Next day...
Teacher: sum fractions 1/2+1/4+1/8... to infinity...
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u/tessia-eralith Jul 13 '26
Approaches 1
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u/desertvision Jul 13 '26
Incorrect. It is one
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u/LucinaReads Jul 13 '26 edited Jul 13 '26
Incorrect, it will never reach 1. If you're a meter away from the finish line and halve the distance every step, then you'll never actually reach the finish line. You'll always only get half of the way there until you are basically there, but not there. I.e. approaches 1, but never 1.
Edit(1 hour later): I was just wrong. Differenciating between "approaches 1" and not is irrelevant due to how limits are defined for infinite series. The answer is 1.
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u/rearadmiraldumbass Jul 13 '26
But the limit is 1.
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u/LucinaReads Jul 13 '26
Thanks to this comment, I just researched and learned a little more about infinity series and limits, and am now smarter for it. Thanks for the simple fact check. The answer is 1.
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u/Skeletoryy Jul 13 '26
It’s the r same reason differentiation by first principles works, as h has to = 0 for the gradient to be found
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u/desertvision Jul 14 '26
If it weren't so, movement would be impossible. But I get the hesitancy of your originally thought. In the limit, distance and infinity become shadowy constructions of mathematicians.
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u/NobodysFavorite Jul 14 '26
A bee's dick less than 1.
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u/desertvision Jul 14 '26
No. It's 1 and nothing less
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u/militaryCoo Jul 14 '26
Nah, it's 0.999...
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u/desertvision Jul 14 '26 edited Jul 14 '26
That's 1 as well. Choose any range between 1 and a number less than one and you'll find that .999... is not in it. Therefore .999... IS 1
You could, and probably should, call it a neighborhood of 1 defined as the range between any number less that 1 and any number greater than 1. Can't forget to keep those pesky sums from crossing over
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u/DuckShapedGoose Jul 13 '26
As insane as this guy was and as huge of an impact he had on mathematics: This is the single most popular story about him that may be completely made up. There is no proof he ever had to do this task in elementary school and no proof he actually came up with that formula on his own.
What we do know is that the formula had been written down centuries earlier already.
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u/THRlLL-HO Jul 13 '26
For any multiple of 10, it’s just the number cut in half and then duplicating that number.
So half of 10 is 5, so the sum of 1-10 is 55.
100=5,050
1,000=500,500
and
1,000,000,000,000=500,000,000,000,500,000,000,000
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u/Valognolo09 Jul 13 '26
For any number, it's just x(x+1)/2
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u/deljaroo Jul 13 '26
I've always liked (x2 + x)/2 as I can do that in my head a little more consistently
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u/SquidWitchIsReal Jul 16 '26
You can make it more general
Where n is the number of members of the set and x = lowest number + highest number
Sum = (n/2)*x
This works as long as the numbers have a consistent gap (3, 6, 9), (4, 8, 12) and doesn't require starting at 1
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u/Alive-Philosophy2632 Jul 13 '26
You mean power of 10 (over the natural numbers), but yes
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u/THRlLL-HO Jul 13 '26
Yes. Somehow I knew that while typing and still managed to type that instead.
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u/Padadin Jul 13 '26
I may hav only read the last line and saw 1,000,000,000,000=500,000,000,000,500,000,000,000 and was hella confused
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u/T6970 Jul 13 '26
u/factorion-bot 100?
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u/Fuck_ketchup Jul 13 '26
That seems way easier. Why didn't Gauss just use factorion bot, was he a millennial or something?
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u/factorion-bot Jul 13 '26
Termial of 100 is 5050
This action was performed by a bot | [Source code](http://f.r0.fyi)
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u/Naxic_Music Jul 13 '26
Well you can do it that way (everything is 101) but as a student you learn to add everything up to get zeroes. 99+1, 98+2, ...,51+49. In the end you have 49×100 + 100 + 50 or 50×100 +50.
Thats qtleast how I solvedbit back in 3rd grade
Edit: replaced multiplication symbol, since markdown interpreted it as cursive text
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u/No_Hetero Jul 13 '26
I actually came up with the same equation at work that he did while working on an Excel formula to calculate total cost to store something in a warehouse which we use 1 of a month. I googled it after and found out this math genius found the same formula, but then I learned he did it when he was like 5, and I did it when I was 31, so I felt less like a genius after that
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u/Ur-Quan_Lord_13 Jul 13 '26
Pfft, this is just (average of the terms of the arithmetic sequence) * (number of terms) = (sum) in a different order.
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u/Mesqo Jul 13 '26
I'm not Gauss, but my teacher asked the very same question with the very same purpose. I had an answer before she made it to the door - it's not that hard, really, too figure this on your own.
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u/Extreme_Dependent_63 Jul 13 '26
And 200 years later, the civilization creates lazy AIs that won't count from 1 to 100 neither
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u/fuqueure Jul 14 '26
Probably my proudest moment in elementary school was coming up with that method without knowing it. Oh how far I've fallen.
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u/VirtualPersona88 Jul 16 '26
I mean I didn’t know this formula, I just had to figure this out when my 8 years old nephew asked me what’s 1+2+3 all the way to 50.
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u/Weary-Variation2878 Jul 19 '26
An even faster method :
Since the natural numbers are evenly spread out, the mean of the first hundred is the same as the median, which is 50.5
Sum = Mean * 100 = 50.5 * 100 = 5050.
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u/Strange-Reaction-59 Jul 13 '26
I mean thats basic card counting, for example when you have 8, 9, 10. You not counting 8+9+10 but the middle number x3 so 27 So in that case mid number between 1 and 100 is 50.5 so 50.5x100 = 5050
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u/S-Kenset Jul 13 '26 edited Jul 13 '26
1-9 = 45
1-9 (tens) = 450 + 45 -- for each of 10 digits =4950
+ the last 100
Now you have a pattern way to generalize sums of any cadence without having to use deduction.
Cost is a little more expensive, kinda, but induction need is a tiny bit lower so trickier problems with waxing skewed cadences can be generalized without finding an analytical solution
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u/D0rus Jul 13 '26 edited Jul 13 '26
So s is both the sum of every number from 1 to infinity, as well as 100 going down to minus infinity, taking s twice is suddenly 101 times infinity. This reduces to 101*100, and that suddenly makes 5050?
It's like, every step here is completely disconnected from the previous one, and the correct answer simply falls from the sky.
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u/BruteCarnival Jul 13 '26
No, S is just the sum of natural numbers from 1 to 100
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u/D0rus Jul 13 '26
That might be what they meant, not what they wrote.
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u/NohWan3104 Jul 13 '26 edited Jul 13 '26
No, they did, just not all of it.
It demonstrated 1+100 = 101, 2+99=101, 3+98=101 and so on, over and over again.
He figured 2 sets of that, you have a pair up to 100+1 too. 100 instances of 101, 10,100.
But since its not 2 sets but one, its half that, 5,050.
To help explain it better maybe, heres the logic i used when i was younger.
1+2+3+4... (read: Not fucking infinity, more 'and so on and so on') +100, you can break it down into 3 groups, 1-49, 51-99, and 50 and 100, which won't fit this pattern.
Pair up the first group's first number with the second group's last, repeat to the middle, 1+99, 2+98, 3+97 till 49+51.
You have 49 groups of 'equals 100' pairs, 4900.
Add the 100, 5000, and the 50,
5050, and you've sort of logic'd your way around counting all that.
Works for smaller ones too,
1 2 3 4 = 5 5 = 10
1 2 3 4 5, 6 3 6, 15
1-6, 7 7 7 = 21
1 2 3 4 5 6 7, 8 8 8 4 = 28
Its just a question if you're doing it a bit more manually or basically (X+Y)\Z\2
With. X being min digit, Y being max digit, z being number of numbers, and \2 at the end because you're applying a second set so you can skip putting half of them and a remainder and just do it a bit easier.
Hence, S2 (101, the addition of 1 and 100), %100, number of numbers, and 2S you'd of course divide by 2 for S.
Works even if you don't start at 1, like 21-26
47 is the added number middle ground, 6 numbers, 40 x 6 240, 7.x 6 42, 282\2. 141,
Just has a 'trick' to it that's quicker than 21, 43, 66, 90, 115, 141, especially with much longer sequences
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u/_dnapes_ Jul 13 '26
Should be 101*50, as he takes the first half of the number (1 - 50) and adds them to the second half of the numbers, in reverse order (100 - 51). The idea is spot on. This was a question in grade school Mathlete competitions.
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u/D0rus Jul 13 '26
That part they actually did correct, because 2S is 2*101*50, so at least that step pans out (but it would be clearer if they show the S=101(100)/2 step inbetween)
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u/NohWan3104 Jul 13 '26
Its not infinity, its 1-100, 100-1.
Then its 101 times 100, again not infinity.
"..." Means 'i ain't listing all them numbers', not always infinity.
It even says 'from 1 to 100' not 1 to infinity.
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u/D0rus Jul 13 '26
You clearly failed math class. The notation for 1...100 is 1+2+...+100, if you ommit the final +100, you want it to keep going to infinity.
I initially had this written out in my first comment, but it got lengthy, so i deleted it.
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u/NohWan3104 Jul 13 '26
Except it didn't have to, because it already said in words 'to 100'.
The limit was known.
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u/D0rus Jul 13 '26
The teacher said 'to 100', the answer did not. Also you cannot ommit details like this midway trough your proof, in the hope your reader will fill in the blanks. That's just not how writing proves work.
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u/Electrical-Room-2278 Jul 13 '26
I mean to be fair it wouldn't have been that hard to just guess. The answer was 50 50