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Difference between revisions of "User:Colball"
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PROOF 3: We proceed using induction. If <math>n = 1</math>, then we have <math>1+0=1^2</math>. Now assume that <math>n</math> works. We prove that <math>n+1</math> works. We add a <math>2n+1</math> on both sides, such that the left side becomes <math>1+2+\cdots + (n+1)+1+2+\cdots + n = n^2 + 2n + 1 = (n+1)^2</math> and we are done with the third proof.
 
PROOF 3: We proceed using induction. If <math>n = 1</math>, then we have <math>1+0=1^2</math>. Now assume that <math>n</math> works. We prove that <math>n+1</math> works. We add a <math>2n+1</math> on both sides, such that the left side becomes <math>1+2+\cdots + (n+1)+1+2+\cdots + n = n^2 + 2n + 1 = (n+1)^2</math> and we are done with the third proof.
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Now I will talk about myself:
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Age: <math>8<x<17</math>
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Joined AoPS: Check my Profile (it's my blog)
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FTW rating: Bad
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Overall alumus level: 8ish
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Gender: Can't tell you, (sorry)
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Goals: Make longest AoPS wiki user page.
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Goals Report: Almost.
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Location: Tucson, AZ
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Brothers username: kix
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Post count: 58, (it would br more if my mom )

Revision as of 14:20, 15 June 2019

Hi, I'm Colball.

You might know me as the founder of AoPS times or the founder of The Cool. The Amazing. The Poll Forum. Anyway, I am going to talk about one of my favorite theorems. It says that $1+2+3+...+n+1+2+3...+(n-1)=n^2$. And here are three proofs:

PROOF 1: $1+2+3+...+n+1+2+3...+(n-1)=n^2$, Hence $\frac{n(n+1)}{2}+\frac{n(n+1)}{2}=n^2$. If you dont get that go to words.Conbine the fractions you get $\frac{n(n+1)+n(n-1)}{2}$. Then Multiply: $\frac{n^2+n+n^2-n}{2}$. Finnaly the $n$'s in the numorator cancel leaving us with $\frac{n^2+n^2}{2}=n^2$. I think you can finish the proof from there.


PROOF 2: The $1+2+\cdots+n$ part refers to an $n$ by $n$ square cut by its diagonal and includes all the squares on the diagonal. The $1+2+\cdots+ n-1$ part refers to an $n$ by $n$ square cut by its diagonal but doesn't include the squares on the diagonal. Putting these together gives us a $n$ by $n$ square.


PROOF 3: We proceed using induction. If $n = 1$, then we have $1+0=1^2$. Now assume that $n$ works. We prove that $n+1$ works. We add a $2n+1$ on both sides, such that the left side becomes $1+2+\cdots + (n+1)+1+2+\cdots + n = n^2 + 2n + 1 = (n+1)^2$ and we are done with the third proof.

Now I will talk about myself: Age: $8<x<17$ Joined AoPS: Check my Profile (it's my blog) FTW rating: Bad Overall alumus level: 8ish Gender: Can't tell you, (sorry) Goals: Make longest AoPS wiki user page. Goals Report: Almost. Location: Tucson, AZ Brothers username: kix Post count: 58, (it would br more if my mom )

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