Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Sums and Perfect Sqares"
(Propose for deletion)
 
Line 1: Line 1:
−
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 theorem 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 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 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.
+
==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.
  
−
 
+
{{delete|unrelated title, unnecessary page}}
−
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.
 
−
 
 
−
 
 
−
Thank you,
 
−
Colball.
 

Latest revision as of 12:47, 19 February 2025

Here are many proofs for the theorem 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.

This article has been proposed for deletion. The reason given is: unrelated title, unnecessary page.

Sysops: Before deleting this article, please check the article discussion pages and history.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Sums_and_Perfect_Sqares&oldid=243215"