User:Colball

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.

Page

Toolbox

Search

User:Colball