Sums and Perfect Sqares

Here are many proofs for the Theory that $1+2+3+...+n+1+2+3...+(n-1)=n^2$

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.

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Sums and Perfect Sqares