- #1

- 46

- 0

## Main Question or Discussion Point

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 mid-point of the leg of the left triangle and drew a line connecting with the mid-point of the hypotenuse, Then with that smaller triangle I just created, I found the mid point of the new leg and the mid-point 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.

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 mid-point of the leg of the left triangle and drew a line connecting with the mid-point of the hypotenuse, Then with that smaller triangle I just created, I found the mid point of the new leg and the mid-point 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.