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> | ||
− | 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 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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+ | Thank you, | ||
+ | Colball. |
Latest revision as of 11:54, 15 June 2019
Here are many proofs for the Theory that
PROOF 1: , Hence
. If you dont get that go to words.Conbine the fractions you get
. Then Multiply:
. Finnaly the
's in the numorator cancel leaving us with
. I think you can finish the proof from there.
PROOF 2: The part refers to an
by
square cut by its diagonal and includes all the squares on the diagonal. The
part refers to an
by
square cut by its diagonal but doesn't include the squares on the diagonal. Putting these together gives us a
by
square.
PROOF 3: We proceed using induction. If , then we have
. Now assume that
works. We prove that
works. We add a
on both sides, such that the left side becomes
and we are done with the third proof.
Thank you,
Colball.