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| − | Hi, I'm Colball.
| + | My password is nothing. (The riddle) |
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| − | 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 <math>1+2+3+...+n+1+2+3...+(n-1)=n^2</math>. And here are three proofs:
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| − | PROOF 1: <math>1+2+3+...+n+1+2+3...+(n-1)=n^2</math>, Hence <math>\frac{n(n+1)}{2}+\frac{n(n+1)}{2}=n^2</math>. If you dont get that go to words.Conbine the fractions you get <math>\frac{n(n+1)+n(n-1)}{2}</math>. Then Multiply: <math>\frac{n^2+n+n^2-n}{2}</math>. Finnaly the <math>n</math>'s in the numorator cancel leaving us with <math>\frac{n^2+n^2}{2}=n^2</math>. I think you can finish the proof from there.
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| − | PROOF 2: The <math>1+2+\cdots+n</math> part refers to an <math>n</math> by <math>n</math> square cut by its diagonal and includes all the squares on the diagonal. The <math>1+2+\cdots+ n-1</math> part refers to an <math>n</math> by <math>n</math> square cut by its diagonal but doesn't include the squares on the diagonal. Putting these together gives us a <math>n</math> by <math>n</math> square.
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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.
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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 )
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Latest revision as of 11:50, 29 May 2020
My password is nothing. (The riddle)
