
#1
Nov709, 07:11 PM

P: 47

I was bored one day, so my father told me to add all the numbers between one and 100 with these conditions:
1. Do not look on the internet. 2. do not add all the numbers up like 1+2+3+4+5+6+7.... So I set out on my journey looking at all possibilities, then I gave up. but I never looked on the internet, so many days latter I was in chemistry and I had an epiphany while doing electron configuration. I divided a square along it's diagonal, and shaded in the right side. Then I found the midpoint of the leg of the left triangle and drew a line connecting with the midpoint of the hypotenuse, Then with that smaller triangle I just created, I found the mid point of the new leg and the midpoint of the new hypotenuses. I shaded in the tiniest triangle and began creating my equation. [tex]\frac{n(n)+ 1/2(n)}{2}[/tex] which is equivalent to: [tex]\sum[/tex][tex]^{n}_{n=1}[/tex]n So when you add up the area of the small triangle and the large triangle it creates the sum all all numbers between 1 and the length of one of your sides. The cool part about it is I think when you find the area of what is left, (not shaded) it is the sum of all numbers between 1 and 1 less than the length of one of your sides. The reason I share this is for you to check if I am right, before I present it to my dad :D. 



#2
Nov709, 07:41 PM

P: 1,623

If what you've written out is correct, the formula that you found isn't quite right. You can easily verify this by choosing pretty much any positive integer [itex]n[/itex]. Since it sounds like you're trying to come up with this independently, I won't give you any hints. Good luck! :)




#3
Nov709, 08:31 PM

P: 47

[tex]\frac{n(n)}{2}[/tex] +1/2n
:3 simple arrangement mistake, the one I entered was from memory, I was sure I was correct but apparently not, so I looked in my notebooks. 



#4
Nov709, 08:59 PM

P: 1,623

Adding all the numbers between one and 100
That formula is correct! Note that it's more conventional to write the formula for the sum of the first [itex]n[/itex] natural numbers as:
[tex]\sum_{i=1}^ni = \frac{n(n+1)}{2}[/tex] Good job! 



#5
Nov709, 09:14 PM

P: 47

[tex]\sum^{n}_{i=1}i[/tex]
what does this mean xD edit: or better, how do the imaginary numbers fit into it :x 



#6
Nov709, 09:23 PM

P: 878





#7
Nov709, 09:26 PM

HW Helper
P: 3,436

Haha it has nothing to do with imaginary numbers
It's just saying that the first term of i is 1 and you keep summing up this variable i for all numbers until n. The same way [tex]\sum^n_{k=1}k=\frac{n(n+1)}{2}[/tex] I like how you've figured out this problem Funny how your epiphany sent you down a much more complicated road to solve this problem for all n integers! Rather than just the first 100. I could think of a much simpler way to solve what your father asked for, but you should show him your method instead 



#8
Nov709, 09:34 PM

P: 47

Oh yes of course xD I know how to do that then...thank you all :D




#9
Nov809, 04:19 AM

Sci Advisor
HW Helper
PF Gold
P: 12,016

Congratulations, yyttr2!
It's fun when you find out something clever like this by yourself, agreed? There's a story about how one of the greatest mathematicians in history, Carl Friedrich Gauss, solved the same problem as you: At this time, he was 7, and his teacher was bored with teaching, and said to his pupils they should add together the numbers from 1 to 100 (thus, he thought he would gain a bit of rest). He was very annoyed when little Carl Friedrich came up to him after only a couple of minutes with the solution: Gauss organized his sum as follows: (1+100)+(2+99)+(3+98) and so on. Thus, he has fifty terms, each equalling 101, so that the sum must be: 50*101=5050. 



#10
Nov809, 11:13 AM

P: 105




#11
Nov909, 09:06 PM

P: 47

It has already been stated by someone else GRRR!




#12
Nov909, 09:39 PM

P: 1,623

Are you upset that someone else found the formula for the sum of the first [itex]n[/itex] postive integers before you did? If you are, that's silly! As a young student of mathematics most, if not all, of your ideas will probably already have been stated by someone else. Should that diminish the significance that you derived that formula on your own? Not in the slightest!




#13
Nov1109, 06:58 PM

P: 13

Haha.. I felt similar sensations when I proved several of the Taylor Polynomials =P.
Congratulations! 


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