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Difference between revisions of "Sums and Perfect Sqares"
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Here are many proofs for the Theory that <math>1+2+3+...+n+1+2+3...+(n-1)=n^2</math>
 
Here are many proofs for the Theory that <math>1+2+3+...+n+1+2+3...+(n-1)=n^2</math>
  
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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 ''Proof without words''. If you 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 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.
  
  
 
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.
 
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.

Revision as of 13:39, 14 June 2019

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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