Sums and Perfect Sqares

Revision as of 12:51, 14 June 2019 by Colball (talk | contribs)

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.


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.


Math is like art in many ways and people sometimes make a hobby of proof after proof after proof. However, these proofs are mostly from the Pythagorean Theorem. This theorem already has more proofs then needed so mathematicians should make a hobby of making proofs for theorems like these.

Thank you, Colball