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Playing with squares

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3

$begingroup$

Extending from particular examples I've found that $$n^2=sum_i=1^i=n-1 2, i+n$$
this is that for any square of side $n$ the area can be calculated in a simple way.

Example

For a square of side $7$, the result is: $2×1+2×2+2×3+cdots + 7=49$

Question

Is there any way to prove this generally true, is there more than one way?Can you show at least one?

square-numbers

share|cite|improve this question

edited May 21 at 11:43

Thomas Shelby

5,0722728

asked May 21 at 11:25

santimirandarpsantimirandarp

253111

$endgroup$

add a comment | 

3

$begingroup$

Extending from particular examples I've found that $$n^2=sum_i=1^i=n-1 2, i+n$$
this is that for any square of side $n$ the area can be calculated in a simple way.

Example

For a square of side $7$, the result is: $2×1+2×2+2×3+cdots + 7=49$

Question

Is there any way to prove this generally true, is there more than one way?Can you show at least one?

square-numbers

share|cite|improve this question

edited May 21 at 11:43

Thomas Shelby

5,0722728

asked May 21 at 11:25

santimirandarpsantimirandarp

253111

$endgroup$

add a comment | 

$begingroup$

Extending from particular examples I've found that $$n^2=sum_i=1^i=n-1 2, i+n$$
this is that for any square of side $n$ the area can be calculated in a simple way.

Example

For a square of side $7$, the result is: $2×1+2×2+2×3+cdots + 7=49$

Question

Is there any way to prove this generally true, is there more than one way?Can you show at least one?

square-numbers

share|cite|improve this question

edited May 21 at 11:43

Thomas Shelby

5,0722728

asked May 21 at 11:25

santimirandarpsantimirandarp

253111

$endgroup$

share|cite|improve this question

edited May 21 at 11:43

Thomas Shelby

5,0722728

asked May 21 at 11:25

santimirandarpsantimirandarp

253111

share|cite|improve this question

edited May 21 at 11:43

Thomas Shelby

5,0722728

asked May 21 at 11:25

santimirandarpsantimirandarp

253111

edited May 21 at 11:43

Thomas Shelby

5,0722728

edited May 21 at 11:43

Thomas Shelby

5,0722728

asked May 21 at 11:25

santimirandarpsantimirandarp

253111

asked May 21 at 11:25

santimirandarpsantimirandarp

253111

4 Answers
4

active

oldest

votes

6

$begingroup$

A visual proof. A variant of the one found in the book "Proofs Without Words"

share|cite|improve this answer

answered May 21 at 11:45

b00n heTb00n heT

10.8k12336

$endgroup$

add a comment | 

3

$begingroup$

Permuting your equation, i.e. moving $n$ to the left, and dividing by $2$, yields the equivalent equation

$$fracn(n-1)2=sum_i=1^n-1 i$$ which is a very well known equation that can easily be proven by induction. Or just look at how Gauss did it when he was a child.

share|cite|improve this answer

answered May 21 at 11:27

5xum5xum

94.2k498164

$endgroup$

add a comment | 

3

$begingroup$

$$sum_i=1^i=n-1 2, i,+n=2left(dfrac n (n-1)2right)+n=n^2$$

share|cite|improve this answer

answered May 21 at 11:30

Thomas ShelbyThomas Shelby

5,0722728

$endgroup$

add a comment | 

3

$begingroup$

Note that the sum can be written as a telescopic sum:
$$beginalignsum_i=1^n-1 2 i+n&=sum_i=1^n-1 (2 i+1)+1=sum_i=1^n-1 ((i+1)^2-i^2)+1\
&=(n^2-1)+1=n^2.
endalign$$
Similarly, we can show
$$sum_i=1^n-1 (3i^2+3i)+n=n^3quadtextorquad sum_i=1^n-1 (4i^3+6i^2+4i)+n=n^4.
$$

share|cite|improve this answer

edited May 22 at 5:32

answered May 21 at 11:36

Robert ZRobert Z

103k1073147

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  wow, it is just impressive. there is something strange in the first equality, though...
  $endgroup$
  – santimirandarp
  May 21 at 11:40
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @santimirandarp Yes, you are right. I fixed it.
  $endgroup$
  – Robert Z
  May 21 at 11:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice! Now the last number appears correctly. But I found the (2i + 1) difficult notation, do you? Maybe it's just a matter of habit...
  $endgroup$
  – santimirandarp
  May 21 at 11:50
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $sum_i=1^n-11$ means the sum of $n-1$ times the number $1$, that is $n-1$.
  $endgroup$
  – Robert Z
  May 21 at 12:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, so the mistake is supposing it needs an argument like $i$. Don't know where it comes from. Thanks again.
  $endgroup$
  – santimirandarp
  May 21 at 12:02
  

  
  
  
  

 | 
show 1 more comment

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6

$begingroup$

A visual proof. A variant of the one found in the book "Proofs Without Words"

share|cite|improve this answer

answered May 21 at 11:45

b00n heTb00n heT

10.8k12336

$endgroup$

add a comment | 

6

$begingroup$

A visual proof. A variant of the one found in the book "Proofs Without Words"

share|cite|improve this answer

answered May 21 at 11:45

b00n heTb00n heT

10.8k12336

$endgroup$

add a comment | 

$begingroup$

A visual proof. A variant of the one found in the book "Proofs Without Words"

share|cite|improve this answer

answered May 21 at 11:45

b00n heTb00n heT

10.8k12336

$endgroup$

share|cite|improve this answer

answered May 21 at 11:45

b00n heTb00n heT

10.8k12336

answered May 21 at 11:45

b00n heTb00n heT

10.8k12336

answered May 21 at 11:45

b00n heTb00n heT

10.8k12336

3

$begingroup$

Permuting your equation, i.e. moving $n$ to the left, and dividing by $2$, yields the equivalent equation

$$fracn(n-1)2=sum_i=1^n-1 i$$ which is a very well known equation that can easily be proven by induction. Or just look at how Gauss did it when he was a child.

share|cite|improve this answer

answered May 21 at 11:27

5xum5xum

94.2k498164

$endgroup$

add a comment | 

3

$begingroup$

Permuting your equation, i.e. moving $n$ to the left, and dividing by $2$, yields the equivalent equation

$$fracn(n-1)2=sum_i=1^n-1 i$$ which is a very well known equation that can easily be proven by induction. Or just look at how Gauss did it when he was a child.

share|cite|improve this answer

answered May 21 at 11:27

5xum5xum

94.2k498164

$endgroup$

add a comment | 

$begingroup$

Permuting your equation, i.e. moving $n$ to the left, and dividing by $2$, yields the equivalent equation

$$fracn(n-1)2=sum_i=1^n-1 i$$ which is a very well known equation that can easily be proven by induction. Or just look at how Gauss did it when he was a child.

share|cite|improve this answer

answered May 21 at 11:27

5xum5xum

94.2k498164

$endgroup$

share|cite|improve this answer

answered May 21 at 11:27

5xum5xum

94.2k498164

answered May 21 at 11:27

5xum5xum

94.2k498164

answered May 21 at 11:27

5xum5xum

94.2k498164

3

$begingroup$

$$sum_i=1^i=n-1 2, i,+n=2left(dfrac n (n-1)2right)+n=n^2$$

share|cite|improve this answer

answered May 21 at 11:30

Thomas ShelbyThomas Shelby

5,0722728

$endgroup$

add a comment | 

3

$begingroup$

$$sum_i=1^i=n-1 2, i,+n=2left(dfrac n (n-1)2right)+n=n^2$$

share|cite|improve this answer

answered May 21 at 11:30

Thomas ShelbyThomas Shelby

5,0722728

$endgroup$

add a comment | 

$begingroup$

$$sum_i=1^i=n-1 2, i,+n=2left(dfrac n (n-1)2right)+n=n^2$$

share|cite|improve this answer

answered May 21 at 11:30

Thomas ShelbyThomas Shelby

5,0722728

$endgroup$

share|cite|improve this answer

answered May 21 at 11:30

Thomas ShelbyThomas Shelby

5,0722728

answered May 21 at 11:30

Thomas ShelbyThomas Shelby

5,0722728

answered May 21 at 11:30

Thomas ShelbyThomas Shelby

5,0722728

3

$begingroup$

Note that the sum can be written as a telescopic sum:
$$beginalignsum_i=1^n-1 2 i+n&=sum_i=1^n-1 (2 i+1)+1=sum_i=1^n-1 ((i+1)^2-i^2)+1\
&=(n^2-1)+1=n^2.
endalign$$
Similarly, we can show
$$sum_i=1^n-1 (3i^2+3i)+n=n^3quadtextorquad sum_i=1^n-1 (4i^3+6i^2+4i)+n=n^4.
$$

share|cite|improve this answer

edited May 22 at 5:32

answered May 21 at 11:36

Robert ZRobert Z

103k1073147

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  wow, it is just impressive. there is something strange in the first equality, though...
  $endgroup$
  – santimirandarp
  May 21 at 11:40
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @santimirandarp Yes, you are right. I fixed it.
  $endgroup$
  – Robert Z
  May 21 at 11:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice! Now the last number appears correctly. But I found the (2i + 1) difficult notation, do you? Maybe it's just a matter of habit...
  $endgroup$
  – santimirandarp
  May 21 at 11:50
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $sum_i=1^n-11$ means the sum of $n-1$ times the number $1$, that is $n-1$.
  $endgroup$
  – Robert Z
  May 21 at 12:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, so the mistake is supposing it needs an argument like $i$. Don't know where it comes from. Thanks again.
  $endgroup$
  – santimirandarp
  May 21 at 12:02
  

  
  
  
  

 | 
show 1 more comment

3

$begingroup$

Note that the sum can be written as a telescopic sum:
$$beginalignsum_i=1^n-1 2 i+n&=sum_i=1^n-1 (2 i+1)+1=sum_i=1^n-1 ((i+1)^2-i^2)+1\
&=(n^2-1)+1=n^2.
endalign$$
Similarly, we can show
$$sum_i=1^n-1 (3i^2+3i)+n=n^3quadtextorquad sum_i=1^n-1 (4i^3+6i^2+4i)+n=n^4.
$$

share|cite|improve this answer

edited May 22 at 5:32

answered May 21 at 11:36

Robert ZRobert Z

103k1073147

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  wow, it is just impressive. there is something strange in the first equality, though...
  $endgroup$
  – santimirandarp
  May 21 at 11:40
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @santimirandarp Yes, you are right. I fixed it.
  $endgroup$
  – Robert Z
  May 21 at 11:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice! Now the last number appears correctly. But I found the (2i + 1) difficult notation, do you? Maybe it's just a matter of habit...
  $endgroup$
  – santimirandarp
  May 21 at 11:50
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $sum_i=1^n-11$ means the sum of $n-1$ times the number $1$, that is $n-1$.
  $endgroup$
  – Robert Z
  May 21 at 12:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, so the mistake is supposing it needs an argument like $i$. Don't know where it comes from. Thanks again.
  $endgroup$
  – santimirandarp
  May 21 at 12:02
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Note that the sum can be written as a telescopic sum:
$$beginalignsum_i=1^n-1 2 i+n&=sum_i=1^n-1 (2 i+1)+1=sum_i=1^n-1 ((i+1)^2-i^2)+1\
&=(n^2-1)+1=n^2.
endalign$$
Similarly, we can show
$$sum_i=1^n-1 (3i^2+3i)+n=n^3quadtextorquad sum_i=1^n-1 (4i^3+6i^2+4i)+n=n^4.
$$

share|cite|improve this answer

edited May 22 at 5:32

answered May 21 at 11:36

Robert ZRobert Z

103k1073147

$endgroup$

share|cite|improve this answer

edited May 22 at 5:32

answered May 21 at 11:36

Robert ZRobert Z

103k1073147

answered May 21 at 11:36

Robert ZRobert Z

103k1073147

answered May 21 at 11:36

Robert ZRobert Z

103k1073147