Understanding some proofs-without-words for sums of consecutive numbers, consecutive squares, consecutive odd...
$begingroup$
I understand how to derive the formulas for sum of squares, consecutive squares, consecutive cubes, and sum of consecutive odd numbers but I don't understand the visual proofs for them.
For the second and third images, I am completely lost.
For the first one I can see that there are $(n+1)$ columns and $n$ rows. I'm assuming that the grey are even and that the white are odd or vice versa? So in order to have an even amount of odds and evens you must divide by two?
How can I create an image for the sum of consecutive odd numbers ($1+3+5+...(2n-1)^2 = n^2$)
algebra-precalculus sums-of-squares
edited Jan 21 at 9:36
Blue
48.6k870154
asked Jan 21 at 8:55
user8358234user8358234
342110
$endgroup$
|
show 1 more comment
$begingroup$
I understand how to derive the formulas for sum of squares, consecutive squares, consecutive cubes, and sum of consecutive odd numbers but I don't understand the visual proofs for them.
For the second and third images, I am completely lost.
For the first one I can see that there are $(n+1)$ columns and $n$ rows. I'm assuming that the grey are even and that the white are odd or vice versa? So in order to have an even amount of odds and evens you must divide by two?
How can I create an image for the sum of consecutive odd numbers ($1+3+5+...(2n-1)^2 = n^2$)
algebra-precalculus sums-of-squares
edited Jan 21 at 9:36
Blue
48.6k870154
asked Jan 21 at 8:55
user8358234user8358234
342110
$endgroup$
$begingroup$
In the first image, the white squares represent the terms of the sum $1+2+3+cdotcdotcdot+n$. Each square can be assumed to have unit area. The first row has $1$ white square, the second $2$ and so on. The sum is then merely the area of the white portion of the image, which is half the area of the $ntimes(n+1)$ rectangle
$endgroup$
– Shubham Johri
Jan 21 at 9:00
1
$begingroup$
I believe the second PWoW finds inspiration in the fact that $n^2$ is the sum of the first $n$ odd natural numbers.$$1^2+2^2+3^2+cdotcdotcdot+n^2=1+(1+3)+(1+3+5)+cdotcdotcdot+(1+3+5+cdotcdotcdot+2n-1)\=ntimes1+(n-1)times3+cdotcdotcdot+2times(2n-3)+1times(2n-1)$$
$endgroup$
– Shubham Johri
Jan 21 at 9:06
1
$begingroup$
The question didn't originally ask for a proof of the sum of consecutive odd numbers. (In general, you shouldn't add to a question once you have gotten answers, but be that as it may ...). That the sum of consecutive odd numbers is a square is actually illustrated with the colorful blocks on the left of the second image: note that the bands of color are the consecutive odd numbers.
$endgroup$
– Blue
Jan 21 at 9:40
1
$begingroup$
@Blue thank you. i will refrain from doing that next time
$endgroup$
– user8358234
Jan 21 at 9:41
$begingroup$
@user8358234: Good to know. :) The comment was intended both as (friendly!) advice to you, and as notice to future readers who might wonder why the existing answers don't address the last issue.
$endgroup$
– Blue
Jan 21 at 9:46
|
show 1 more comment
$begingroup$
I understand how to derive the formulas for sum of squares, consecutive squares, consecutive cubes, and sum of consecutive odd numbers but I don't understand the visual proofs for them.
For the second and third images, I am completely lost.
For the first one I can see that there are $(n+1)$ columns and $n$ rows. I'm assuming that the grey are even and that the white are odd or vice versa? So in order to have an even amount of odds and evens you must divide by two?
How can I create an image for the sum of consecutive odd numbers ($1+3+5+...(2n-1)^2 = n^2$)
algebra-precalculus sums-of-squares
edited Jan 21 at 9:36
Blue
48.6k870154
asked Jan 21 at 8:55
user8358234user8358234
342110
$endgroup$
I understand how to derive the formulas for sum of squares, consecutive squares, consecutive cubes, and sum of consecutive odd numbers but I don't understand the visual proofs for them.
For the second and third images, I am completely lost.
For the first one I can see that there are $(n+1)$ columns and $n$ rows. I'm assuming that the grey are even and that the white are odd or vice versa? So in order to have an even amount of odds and evens you must divide by two?
How can I create an image for the sum of consecutive odd numbers ($1+3+5+...(2n-1)^2 = n^2$)
algebra-precalculus sums-of-squares
algebra-precalculus sums-of-squares
edited Jan 21 at 9:36
Blue
48.6k870154
asked Jan 21 at 8:55
user8358234user8358234
342110
edited Jan 21 at 9:36
Blue
48.6k870154
asked Jan 21 at 8:55
user8358234user8358234
342110
edited Jan 21 at 9:36
Blue
48.6k870154
edited Jan 21 at 9:36
Blue
48.6k870154
edited Jan 21 at 9:36
Blue
48.6k870154
48.6k870154
asked Jan 21 at 8:55
user8358234user8358234
342110
asked Jan 21 at 8:55
user8358234user8358234
342110
asked Jan 21 at 8:55
user8358234user8358234
342110
342110
$begingroup$
In the first image, the white squares represent the terms of the sum $1+2+3+cdotcdotcdot+n$. Each square can be assumed to have unit area. The first row has $1$ white square, the second $2$ and so on. The sum is then merely the area of the white portion of the image, which is half the area of the $ntimes(n+1)$ rectangle
$endgroup$
– Shubham Johri
Jan 21 at 9:00
1
$begingroup$
I believe the second PWoW finds inspiration in the fact that $n^2$ is the sum of the first $n$ odd natural numbers.$$1^2+2^2+3^2+cdotcdotcdot+n^2=1+(1+3)+(1+3+5)+cdotcdotcdot+(1+3+5+cdotcdotcdot+2n-1)\=ntimes1+(n-1)times3+cdotcdotcdot+2times(2n-3)+1times(2n-1)$$
$endgroup$
– Shubham Johri
Jan 21 at 9:06
1
$begingroup$
The question didn't originally ask for a proof of the sum of consecutive odd numbers. (In general, you shouldn't add to a question once you have gotten answers, but be that as it may ...). That the sum of consecutive odd numbers is a square is actually illustrated with the colorful blocks on the left of the second image: note that the bands of color are the consecutive odd numbers.
$endgroup$
– Blue
Jan 21 at 9:40
1
$begingroup$
@Blue thank you. i will refrain from doing that next time
$endgroup$
– user8358234
Jan 21 at 9:41
$begingroup$
@user8358234: Good to know. :) The comment was intended both as (friendly!) advice to you, and as notice to future readers who might wonder why the existing answers don't address the last issue.
$endgroup$
– Blue
Jan 21 at 9:46
|
show 1 more comment
$begingroup$
In the first image, the white squares represent the terms of the sum $1+2+3+cdotcdotcdot+n$. Each square can be assumed to have unit area. The first row has $1$ white square, the second $2$ and so on. The sum is then merely the area of the white portion of the image, which is half the area of the $ntimes(n+1)$ rectangle
$endgroup$
– Shubham Johri
Jan 21 at 9:00
1
$begingroup$
I believe the second PWoW finds inspiration in the fact that $n^2$ is the sum of the first $n$ odd natural numbers.$$1^2+2^2+3^2+cdotcdotcdot+n^2=1+(1+3)+(1+3+5)+cdotcdotcdot+(1+3+5+cdotcdotcdot+2n-1)\=ntimes1+(n-1)times3+cdotcdotcdot+2times(2n-3)+1times(2n-1)$$
$endgroup$
– Shubham Johri
Jan 21 at 9:06
1
$begingroup$
The question didn't originally ask for a proof of the sum of consecutive odd numbers. (In general, you shouldn't add to a question once you have gotten answers, but be that as it may ...). That the sum of consecutive odd numbers is a square is actually illustrated with the colorful blocks on the left of the second image: note that the bands of color are the consecutive odd numbers.
$endgroup$
– Blue
Jan 21 at 9:40
1
$begingroup$
@Blue thank you. i will refrain from doing that next time
$endgroup$
– user8358234
Jan 21 at 9:41
$begingroup$
@user8358234: Good to know. :) The comment was intended both as (friendly!) advice to you, and as notice to future readers who might wonder why the existing answers don't address the last issue.
$endgroup$
– Blue
Jan 21 at 9:46
$begingroup$
In the first image, the white squares represent the terms of the sum $1+2+3+cdotcdotcdot+n$. Each square can be assumed to have unit area. The first row has $1$ white square, the second $2$ and so on. The sum is then merely the area of the white portion of the image, which is half the area of the $ntimes(n+1)$ rectangle
$endgroup$
– Shubham Johri
Jan 21 at 9:00
$begingroup$
In the first image, the white squares represent the terms of the sum $1+2+3+cdotcdotcdot+n$. Each square can be assumed to have unit area. The first row has $1$ white square, the second $2$ and so on. The sum is then merely the area of the white portion of the image, which is half the area of the $ntimes(n+1)$ rectangle
$endgroup$
– Shubham Johri
Jan 21 at 9:00
1
1
$begingroup$
I believe the second PWoW finds inspiration in the fact that $n^2$ is the sum of the first $n$ odd natural numbers.$$1^2+2^2+3^2+cdotcdotcdot+n^2=1+(1+3)+(1+3+5)+cdotcdotcdot+(1+3+5+cdotcdotcdot+2n-1)\=ntimes1+(n-1)times3+cdotcdotcdot+2times(2n-3)+1times(2n-1)$$
$endgroup$
– Shubham Johri
Jan 21 at 9:06
$begingroup$
I believe the second PWoW finds inspiration in the fact that $n^2$ is the sum of the first $n$ odd natural numbers.$$1^2+2^2+3^2+cdotcdotcdot+n^2=1+(1+3)+(1+3+5)+cdotcdotcdot+(1+3+5+cdotcdotcdot+2n-1)\=ntimes1+(n-1)times3+cdotcdotcdot+2times(2n-3)+1times(2n-1)$$
$endgroup$
– Shubham Johri
Jan 21 at 9:06
1
1
$begingroup$
The question didn't originally ask for a proof of the sum of consecutive odd numbers. (In general, you shouldn't add to a question once you have gotten answers, but be that as it may ...). That the sum of consecutive odd numbers is a square is actually illustrated with the colorful blocks on the left of the second image: note that the bands of color are the consecutive odd numbers.
$endgroup$
– Blue
Jan 21 at 9:40
$begingroup$
The question didn't originally ask for a proof of the sum of consecutive odd numbers. (In general, you shouldn't add to a question once you have gotten answers, but be that as it may ...). That the sum of consecutive odd numbers is a square is actually illustrated with the colorful blocks on the left of the second image: note that the bands of color are the consecutive odd numbers.
$endgroup$
– Blue
Jan 21 at 9:40
1
1
$begingroup$
@Blue thank you. i will refrain from doing that next time
$endgroup$
– user8358234
Jan 21 at 9:41
$begingroup$
@Blue thank you. i will refrain from doing that next time
$endgroup$
– user8358234
Jan 21 at 9:41
$begingroup$
@user8358234: Good to know. :) The comment was intended both as (friendly!) advice to you, and as notice to future readers who might wonder why the existing answers don't address the last issue.
$endgroup$
– Blue
Jan 21 at 9:46
$begingroup$
@user8358234: Good to know. :) The comment was intended both as (friendly!) advice to you, and as notice to future readers who might wonder why the existing answers don't address the last issue.
$endgroup$
– Blue
Jan 21 at 9:46
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
The second picture gives a visual proof for the formula
$$3(1^2+2^2+3^2+dots +n^2)=frac{n(n+1)}{2}cdot (2n+1)$$
for $n=5$. The sum of the areas of the $3cdot 5$ squares on the right
$$3(1^2+2^2+3^2+4^2+5^2)$$
is equal to the area of the rectangle on the left with height $1+2+3+4+5=frac{6cdot 5}{2}$ (see the first formula) and base $2cdot 5+1$.
The third picture gives a visual proof for the formula
$$4(1^3+2^3+3^3+dots +n^3)=(n(n+1))^2$$
for $n=6$. Starting from the center and evaluating the areas of each concentric frame, the area of the large square of side $7cdot 6$ is
$$4cdot 1^2+8cdot 2^2+12cdot 3^2+16cdot 4^2+20cdot 5^2+24cdot 6^2\
=4(1^3+2^3+3^3+4^3+5^3+6^3)$$
answered Jan 21 at 9:06
Robert ZRobert Z
98.5k1068139
$endgroup$
$begingroup$
for the second picture why is it doing $3(1^2+..._n^2) = frac{n(n+1)}{2}(2n1)$? First I don't know where this 3 comes from, why it is necessary or if it's just arbitrary and secondly shouldn't the denominator be 6 and not 2?
$endgroup$
– user8358234
Jan 21 at 9:34
$begingroup$
In the given picture, on the right side, we have $3$ copies of each square. Of course $3(1^2+2^2+3^2+dots +n^2)=frac{n(n+1)}{2}cdot (2n+1)$ is equivalent to $(1^2+2^2+3^2+dots +n^2)=frac{n(n+1)(2n+1)}{6}$.
$endgroup$
– Robert Z
Jan 21 at 9:38
$begingroup$
ohh got it, sorry that was silly of me. but why are there three copies of the squares? why is it necessary for this proof?
$endgroup$
– user8358234
Jan 21 at 9:40
$begingroup$
The two copies of white squares fill the white part of the rectangle (left and right). The third copies (with colored stripes) fill the central part of the rectangle.
$endgroup$
– Robert Z
Jan 21 at 9:43
$begingroup$
thank you for such a detailed answer. i have one more question about the third image: if n = 6, then the total amount of squares should be 1764 and that is how many total squares there are in that image right? and why is it that the larger square is 7 x7? why don't we start with a 1 x 1 square and end up with a 6 x 6 square as the largest square?
$endgroup$
– user8358234
Jan 21 at 9:52
|
show 2 more comments
$begingroup$
Sum of naturals
The rectangle has $n$ by $n+1$ cells and contains twice the sum of the numbers from $1$ to $n$. Hence
$$2,(1+2+cdots n)=n(n+1).$$
Sum of perfect squares:
The cells from the three sets of squares (of areas $1$ to $n^2$) are rearranged in a rectangle. The height of the rectangle is the sum of integers from $1$ to $n$, while the width is $2n+1$. Hence
$$3,(1+4+cdots n^2)=frac{n(n+1)}2(2n+1).$$
Sum of perfect cubes:
Every ring contains $4k$ squares of area $k^2$, hence in total four times the sum of the $n$ first cubes. At the same time, they form a square of side twice the sum of integers from $1$ to $n$. Hence
$$4,(1+8+cdots n^3)=(n(n+1))^2.$$
To make the second picture even more readable:
answered Jan 21 at 9:25
Yves DaoustYves Daoust
128k675227
$endgroup$
$begingroup$
so for the perfect squares, the height and width you are stating for right side of the image right? but why is that the case that it is that based off the left side of the image
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– user8358234
Jan 21 at 9:43
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@user8358234: observe how the cells are displaced and see that their number is preserved.
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– Yves Daoust
Jan 21 at 9:44
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@user8358234 It took me a while to see it; the white areas are made up of 5x5, 4x4, 3x3, 2x2 and 1x1 squares. They should have colors too.
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– JollyJoker
Jan 21 at 13:11
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@JollyJoker: the white cubes were put on the left side for some reason. When you understand the reason, you have it.
$endgroup$
– Yves Daoust
Jan 21 at 13:46
add a comment |
$begingroup$
For a diagram of the sum of consecutive odd nos, use the diagram in your sum of squares that shows a red L shape with an inner yellow L shape, with an inner blue L shape with an inner green L shape with a single green square.
answered Jan 21 at 11:13
jsc42jsc42
1
$endgroup$
$begingroup$
Can you please elaborate?
$endgroup$
– Michael Wang
Jan 22 at 0:29
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The second picture gives a visual proof for the formula
$$3(1^2+2^2+3^2+dots +n^2)=frac{n(n+1)}{2}cdot (2n+1)$$
for $n=5$. The sum of the areas of the $3cdot 5$ squares on the right
$$3(1^2+2^2+3^2+4^2+5^2)$$
is equal to the area of the rectangle on the left with height $1+2+3+4+5=frac{6cdot 5}{2}$ (see the first formula) and base $2cdot 5+1$.
The third picture gives a visual proof for the formula
$$4(1^3+2^3+3^3+dots +n^3)=(n(n+1))^2$$
for $n=6$. Starting from the center and evaluating the areas of each concentric frame, the area of the large square of side $7cdot 6$ is
$$4cdot 1^2+8cdot 2^2+12cdot 3^2+16cdot 4^2+20cdot 5^2+24cdot 6^2\
=4(1^3+2^3+3^3+4^3+5^3+6^3)$$
answered Jan 21 at 9:06
Robert ZRobert Z
98.5k1068139
$endgroup$
$begingroup$
for the second picture why is it doing $3(1^2+..._n^2) = frac{n(n+1)}{2}(2n1)$? First I don't know where this 3 comes from, why it is necessary or if it's just arbitrary and secondly shouldn't the denominator be 6 and not 2?
$endgroup$
– user8358234
Jan 21 at 9:34
$begingroup$
In the given picture, on the right side, we have $3$ copies of each square. Of course $3(1^2+2^2+3^2+dots +n^2)=frac{n(n+1)}{2}cdot (2n+1)$ is equivalent to $(1^2+2^2+3^2+dots +n^2)=frac{n(n+1)(2n+1)}{6}$.
$endgroup$
– Robert Z
Jan 21 at 9:38
$begingroup$
ohh got it, sorry that was silly of me. but why are there three copies of the squares? why is it necessary for this proof?
$endgroup$
– user8358234
Jan 21 at 9:40
$begingroup$
The two copies of white squares fill the white part of the rectangle (left and right). The third copies (with colored stripes) fill the central part of the rectangle.
$endgroup$
– Robert Z
Jan 21 at 9:43
$begingroup$
thank you for such a detailed answer. i have one more question about the third image: if n = 6, then the total amount of squares should be 1764 and that is how many total squares there are in that image right? and why is it that the larger square is 7 x7? why don't we start with a 1 x 1 square and end up with a 6 x 6 square as the largest square?
$endgroup$
– user8358234
Jan 21 at 9:52
|
show 2 more comments
$begingroup$
The second picture gives a visual proof for the formula
$$3(1^2+2^2+3^2+dots +n^2)=frac{n(n+1)}{2}cdot (2n+1)$$
for $n=5$. The sum of the areas of the $3cdot 5$ squares on the right
$$3(1^2+2^2+3^2+4^2+5^2)$$
is equal to the area of the rectangle on the left with height $1+2+3+4+5=frac{6cdot 5}{2}$ (see the first formula) and base $2cdot 5+1$.
The third picture gives a visual proof for the formula
$$4(1^3+2^3+3^3+dots +n^3)=(n(n+1))^2$$
for $n=6$. Starting from the center and evaluating the areas of each concentric frame, the area of the large square of side $7cdot 6$ is
$$4cdot 1^2+8cdot 2^2+12cdot 3^2+16cdot 4^2+20cdot 5^2+24cdot 6^2\
=4(1^3+2^3+3^3+4^3+5^3+6^3)$$
answered Jan 21 at 9:06
Robert ZRobert Z
98.5k1068139
$endgroup$
$begingroup$
for the second picture why is it doing $3(1^2+..._n^2) = frac{n(n+1)}{2}(2n1)$? First I don't know where this 3 comes from, why it is necessary or if it's just arbitrary and secondly shouldn't the denominator be 6 and not 2?
$endgroup$
– user8358234
Jan 21 at 9:34
$begingroup$
In the given picture, on the right side, we have $3$ copies of each square. Of course $3(1^2+2^2+3^2+dots +n^2)=frac{n(n+1)}{2}cdot (2n+1)$ is equivalent to $(1^2+2^2+3^2+dots +n^2)=frac{n(n+1)(2n+1)}{6}$.
$endgroup$
– Robert Z
Jan 21 at 9:38
$begingroup$
ohh got it, sorry that was silly of me. but why are there three copies of the squares? why is it necessary for this proof?
$endgroup$
– user8358234
Jan 21 at 9:40
$begingroup$
The two copies of white squares fill the white part of the rectangle (left and right). The third copies (with colored stripes) fill the central part of the rectangle.
$endgroup$
– Robert Z
Jan 21 at 9:43
$begingroup$
thank you for such a detailed answer. i have one more question about the third image: if n = 6, then the total amount of squares should be 1764 and that is how many total squares there are in that image right? and why is it that the larger square is 7 x7? why don't we start with a 1 x 1 square and end up with a 6 x 6 square as the largest square?
$endgroup$
– user8358234
Jan 21 at 9:52
|
show 2 more comments
$begingroup$
The second picture gives a visual proof for the formula
$$3(1^2+2^2+3^2+dots +n^2)=frac{n(n+1)}{2}cdot (2n+1)$$
for $n=5$. The sum of the areas of the $3cdot 5$ squares on the right
$$3(1^2+2^2+3^2+4^2+5^2)$$
is equal to the area of the rectangle on the left with height $1+2+3+4+5=frac{6cdot 5}{2}$ (see the first formula) and base $2cdot 5+1$.
The third picture gives a visual proof for the formula
$$4(1^3+2^3+3^3+dots +n^3)=(n(n+1))^2$$
for $n=6$. Starting from the center and evaluating the areas of each concentric frame, the area of the large square of side $7cdot 6$ is
$$4cdot 1^2+8cdot 2^2+12cdot 3^2+16cdot 4^2+20cdot 5^2+24cdot 6^2\
=4(1^3+2^3+3^3+4^3+5^3+6^3)$$
answered Jan 21 at 9:06
Robert ZRobert Z
98.5k1068139
$endgroup$
The second picture gives a visual proof for the formula
$$3(1^2+2^2+3^2+dots +n^2)=frac{n(n+1)}{2}cdot (2n+1)$$
for $n=5$. The sum of the areas of the $3cdot 5$ squares on the right
$$3(1^2+2^2+3^2+4^2+5^2)$$
is equal to the area of the rectangle on the left with height $1+2+3+4+5=frac{6cdot 5}{2}$ (see the first formula) and base $2cdot 5+1$.
The third picture gives a visual proof for the formula
$$4(1^3+2^3+3^3+dots +n^3)=(n(n+1))^2$$
for $n=6$. Starting from the center and evaluating the areas of each concentric frame, the area of the large square of side $7cdot 6$ is
$$4cdot 1^2+8cdot 2^2+12cdot 3^2+16cdot 4^2+20cdot 5^2+24cdot 6^2\
=4(1^3+2^3+3^3+4^3+5^3+6^3)$$
answered Jan 21 at 9:06
Robert ZRobert Z
98.5k1068139
edited Jan 21 at 9:36
answered Jan 21 at 9:06
Robert ZRobert Z
98.5k1068139
answered Jan 21 at 9:06
Robert ZRobert Z
98.5k1068139
answered Jan 21 at 9:06
Robert ZRobert Z
98.5k1068139
98.5k1068139
$begingroup$
for the second picture why is it doing $3(1^2+..._n^2) = frac{n(n+1)}{2}(2n1)$? First I don't know where this 3 comes from, why it is necessary or if it's just arbitrary and secondly shouldn't the denominator be 6 and not 2?
$endgroup$
– user8358234
Jan 21 at 9:34
$begingroup$
In the given picture, on the right side, we have $3$ copies of each square. Of course $3(1^2+2^2+3^2+dots +n^2)=frac{n(n+1)}{2}cdot (2n+1)$ is equivalent to $(1^2+2^2+3^2+dots +n^2)=frac{n(n+1)(2n+1)}{6}$.
$endgroup$
– Robert Z
Jan 21 at 9:38
$begingroup$
ohh got it, sorry that was silly of me. but why are there three copies of the squares? why is it necessary for this proof?
$endgroup$
– user8358234
Jan 21 at 9:40
$begingroup$
The two copies of white squares fill the white part of the rectangle (left and right). The third copies (with colored stripes) fill the central part of the rectangle.
$endgroup$
– Robert Z
Jan 21 at 9:43
$begingroup$
thank you for such a detailed answer. i have one more question about the third image: if n = 6, then the total amount of squares should be 1764 and that is how many total squares there are in that image right? and why is it that the larger square is 7 x7? why don't we start with a 1 x 1 square and end up with a 6 x 6 square as the largest square?
$endgroup$
– user8358234
Jan 21 at 9:52
|
show 2 more comments
$begingroup$
for the second picture why is it doing $3(1^2+..._n^2) = frac{n(n+1)}{2}(2n1)$? First I don't know where this 3 comes from, why it is necessary or if it's just arbitrary and secondly shouldn't the denominator be 6 and not 2?
$endgroup$
– user8358234
Jan 21 at 9:34
$begingroup$
In the given picture, on the right side, we have $3$ copies of each square. Of course $3(1^2+2^2+3^2+dots +n^2)=frac{n(n+1)}{2}cdot (2n+1)$ is equivalent to $(1^2+2^2+3^2+dots +n^2)=frac{n(n+1)(2n+1)}{6}$.
$endgroup$
– Robert Z
Jan 21 at 9:38
$begingroup$
ohh got it, sorry that was silly of me. but why are there three copies of the squares? why is it necessary for this proof?
$endgroup$
– user8358234
Jan 21 at 9:40
$begingroup$
The two copies of white squares fill the white part of the rectangle (left and right). The third copies (with colored stripes) fill the central part of the rectangle.
$endgroup$
– Robert Z
Jan 21 at 9:43
$begingroup$
thank you for such a detailed answer. i have one more question about the third image: if n = 6, then the total amount of squares should be 1764 and that is how many total squares there are in that image right? and why is it that the larger square is 7 x7? why don't we start with a 1 x 1 square and end up with a 6 x 6 square as the largest square?
$endgroup$
– user8358234
Jan 21 at 9:52
$begingroup$
for the second picture why is it doing $3(1^2+..._n^2) = frac{n(n+1)}{2}(2n1)$? First I don't know where this 3 comes from, why it is necessary or if it's just arbitrary and secondly shouldn't the denominator be 6 and not 2?
$endgroup$
– user8358234
Jan 21 at 9:34
$begingroup$
for the second picture why is it doing $3(1^2+..._n^2) = frac{n(n+1)}{2}(2n1)$? First I don't know where this 3 comes from, why it is necessary or if it's just arbitrary and secondly shouldn't the denominator be 6 and not 2?
$endgroup$
– user8358234
Jan 21 at 9:34
$begingroup$
In the given picture, on the right side, we have $3$ copies of each square. Of course $3(1^2+2^2+3^2+dots +n^2)=frac{n(n+1)}{2}cdot (2n+1)$ is equivalent to $(1^2+2^2+3^2+dots +n^2)=frac{n(n+1)(2n+1)}{6}$.
$endgroup$
– Robert Z
Jan 21 at 9:38
$begingroup$
In the given picture, on the right side, we have $3$ copies of each square. Of course $3(1^2+2^2+3^2+dots +n^2)=frac{n(n+1)}{2}cdot (2n+1)$ is equivalent to $(1^2+2^2+3^2+dots +n^2)=frac{n(n+1)(2n+1)}{6}$.
$endgroup$
– Robert Z
Jan 21 at 9:38
$begingroup$
ohh got it, sorry that was silly of me. but why are there three copies of the squares? why is it necessary for this proof?
$endgroup$
– user8358234
Jan 21 at 9:40
$begingroup$
ohh got it, sorry that was silly of me. but why are there three copies of the squares? why is it necessary for this proof?
$endgroup$
– user8358234
Jan 21 at 9:40
$begingroup$
The two copies of white squares fill the white part of the rectangle (left and right). The third copies (with colored stripes) fill the central part of the rectangle.
$endgroup$
– Robert Z
Jan 21 at 9:43
$begingroup$
The two copies of white squares fill the white part of the rectangle (left and right). The third copies (with colored stripes) fill the central part of the rectangle.
$endgroup$
– Robert Z
Jan 21 at 9:43
$begingroup$
thank you for such a detailed answer. i have one more question about the third image: if n = 6, then the total amount of squares should be 1764 and that is how many total squares there are in that image right? and why is it that the larger square is 7 x7? why don't we start with a 1 x 1 square and end up with a 6 x 6 square as the largest square?
$endgroup$
– user8358234
Jan 21 at 9:52
$begingroup$
thank you for such a detailed answer. i have one more question about the third image: if n = 6, then the total amount of squares should be 1764 and that is how many total squares there are in that image right? and why is it that the larger square is 7 x7? why don't we start with a 1 x 1 square and end up with a 6 x 6 square as the largest square?
$endgroup$
– user8358234
Jan 21 at 9:52
|
show 2 more comments
$begingroup$
Sum of naturals
The rectangle has $n$ by $n+1$ cells and contains twice the sum of the numbers from $1$ to $n$. Hence
$$2,(1+2+cdots n)=n(n+1).$$
Sum of perfect squares:
The cells from the three sets of squares (of areas $1$ to $n^2$) are rearranged in a rectangle. The height of the rectangle is the sum of integers from $1$ to $n$, while the width is $2n+1$. Hence
$$3,(1+4+cdots n^2)=frac{n(n+1)}2(2n+1).$$
Sum of perfect cubes:
Every ring contains $4k$ squares of area $k^2$, hence in total four times the sum of the $n$ first cubes. At the same time, they form a square of side twice the sum of integers from $1$ to $n$. Hence
$$4,(1+8+cdots n^3)=(n(n+1))^2.$$
To make the second picture even more readable:
answered Jan 21 at 9:25
Yves DaoustYves Daoust
128k675227
$endgroup$
$begingroup$
so for the perfect squares, the height and width you are stating for right side of the image right? but why is that the case that it is that based off the left side of the image
$endgroup$
– user8358234
Jan 21 at 9:43
$begingroup$
@user8358234: observe how the cells are displaced and see that their number is preserved.
$endgroup$
– Yves Daoust
Jan 21 at 9:44
$begingroup$
@user8358234 It took me a while to see it; the white areas are made up of 5x5, 4x4, 3x3, 2x2 and 1x1 squares. They should have colors too.
$endgroup$
– JollyJoker
Jan 21 at 13:11
$begingroup$
@JollyJoker: the white cubes were put on the left side for some reason. When you understand the reason, you have it.
$endgroup$
– Yves Daoust
Jan 21 at 13:46
add a comment |
$begingroup$
Sum of naturals
The rectangle has $n$ by $n+1$ cells and contains twice the sum of the numbers from $1$ to $n$. Hence
$$2,(1+2+cdots n)=n(n+1).$$
Sum of perfect squares:
The cells from the three sets of squares (of areas $1$ to $n^2$) are rearranged in a rectangle. The height of the rectangle is the sum of integers from $1$ to $n$, while the width is $2n+1$. Hence
$$3,(1+4+cdots n^2)=frac{n(n+1)}2(2n+1).$$
Sum of perfect cubes:
Every ring contains $4k$ squares of area $k^2$, hence in total four times the sum of the $n$ first cubes. At the same time, they form a square of side twice the sum of integers from $1$ to $n$. Hence
$$4,(1+8+cdots n^3)=(n(n+1))^2.$$
To make the second picture even more readable:
answered Jan 21 at 9:25
Yves DaoustYves Daoust
128k675227
$endgroup$
$begingroup$
so for the perfect squares, the height and width you are stating for right side of the image right? but why is that the case that it is that based off the left side of the image
$endgroup$
– user8358234
Jan 21 at 9:43
$begingroup$
@user8358234: observe how the cells are displaced and see that their number is preserved.
$endgroup$
– Yves Daoust
Jan 21 at 9:44
$begingroup$
@user8358234 It took me a while to see it; the white areas are made up of 5x5, 4x4, 3x3, 2x2 and 1x1 squares. They should have colors too.
$endgroup$
– JollyJoker
Jan 21 at 13:11
$begingroup$
@JollyJoker: the white cubes were put on the left side for some reason. When you understand the reason, you have it.
$endgroup$
– Yves Daoust
Jan 21 at 13:46
add a comment |
$begingroup$
Sum of naturals
The rectangle has $n$ by $n+1$ cells and contains twice the sum of the numbers from $1$ to $n$. Hence
$$2,(1+2+cdots n)=n(n+1).$$
Sum of perfect squares:
The cells from the three sets of squares (of areas $1$ to $n^2$) are rearranged in a rectangle. The height of the rectangle is the sum of integers from $1$ to $n$, while the width is $2n+1$. Hence
$$3,(1+4+cdots n^2)=frac{n(n+1)}2(2n+1).$$
Sum of perfect cubes:
Every ring contains $4k$ squares of area $k^2$, hence in total four times the sum of the $n$ first cubes. At the same time, they form a square of side twice the sum of integers from $1$ to $n$. Hence
$$4,(1+8+cdots n^3)=(n(n+1))^2.$$
To make the second picture even more readable:
answered Jan 21 at 9:25
Yves DaoustYves Daoust
128k675227
$endgroup$
Sum of naturals
The rectangle has $n$ by $n+1$ cells and contains twice the sum of the numbers from $1$ to $n$. Hence
$$2,(1+2+cdots n)=n(n+1).$$
Sum of perfect squares:
The cells from the three sets of squares (of areas $1$ to $n^2$) are rearranged in a rectangle. The height of the rectangle is the sum of integers from $1$ to $n$, while the width is $2n+1$. Hence
$$3,(1+4+cdots n^2)=frac{n(n+1)}2(2n+1).$$
Sum of perfect cubes:
Every ring contains $4k$ squares of area $k^2$, hence in total four times the sum of the $n$ first cubes. At the same time, they form a square of side twice the sum of integers from $1$ to $n$. Hence
$$4,(1+8+cdots n^3)=(n(n+1))^2.$$
To make the second picture even more readable:
answered Jan 21 at 9:25
Yves DaoustYves Daoust
128k675227
edited Jan 21 at 13:53
answered Jan 21 at 9:25
Yves DaoustYves Daoust
128k675227
answered Jan 21 at 9:25
Yves DaoustYves Daoust
128k675227
answered Jan 21 at 9:25
Yves DaoustYves Daoust
128k675227
128k675227
$begingroup$
so for the perfect squares, the height and width you are stating for right side of the image right? but why is that the case that it is that based off the left side of the image
$endgroup$
– user8358234
Jan 21 at 9:43
$begingroup$
@user8358234: observe how the cells are displaced and see that their number is preserved.
$endgroup$
– Yves Daoust
Jan 21 at 9:44
$begingroup$
@user8358234 It took me a while to see it; the white areas are made up of 5x5, 4x4, 3x3, 2x2 and 1x1 squares. They should have colors too.
$endgroup$
– JollyJoker
Jan 21 at 13:11
$begingroup$
@JollyJoker: the white cubes were put on the left side for some reason. When you understand the reason, you have it.
$endgroup$
– Yves Daoust
Jan 21 at 13:46
add a comment |
$begingroup$
so for the perfect squares, the height and width you are stating for right side of the image right? but why is that the case that it is that based off the left side of the image
$endgroup$
– user8358234
Jan 21 at 9:43
$begingroup$
@user8358234: observe how the cells are displaced and see that their number is preserved.
$endgroup$
– Yves Daoust
Jan 21 at 9:44
$begingroup$
@user8358234 It took me a while to see it; the white areas are made up of 5x5, 4x4, 3x3, 2x2 and 1x1 squares. They should have colors too.
$endgroup$
– JollyJoker
Jan 21 at 13:11
$begingroup$
@JollyJoker: the white cubes were put on the left side for some reason. When you understand the reason, you have it.
$endgroup$
– Yves Daoust
Jan 21 at 13:46
$begingroup$
so for the perfect squares, the height and width you are stating for right side of the image right? but why is that the case that it is that based off the left side of the image
$endgroup$
– user8358234
Jan 21 at 9:43
$begingroup$
so for the perfect squares, the height and width you are stating for right side of the image right? but why is that the case that it is that based off the left side of the image
$endgroup$
– user8358234
Jan 21 at 9:43
$begingroup$
@user8358234: observe how the cells are displaced and see that their number is preserved.
$endgroup$
– Yves Daoust
Jan 21 at 9:44
$begingroup$
@user8358234: observe how the cells are displaced and see that their number is preserved.
$endgroup$
– Yves Daoust
Jan 21 at 9:44
$begingroup$
@user8358234 It took me a while to see it; the white areas are made up of 5x5, 4x4, 3x3, 2x2 and 1x1 squares. They should have colors too.
$endgroup$
– JollyJoker
Jan 21 at 13:11
$begingroup$
@user8358234 It took me a while to see it; the white areas are made up of 5x5, 4x4, 3x3, 2x2 and 1x1 squares. They should have colors too.
$endgroup$
– JollyJoker
Jan 21 at 13:11
$begingroup$
@JollyJoker: the white cubes were put on the left side for some reason. When you understand the reason, you have it.
$endgroup$
– Yves Daoust
Jan 21 at 13:46
$begingroup$
@JollyJoker: the white cubes were put on the left side for some reason. When you understand the reason, you have it.
$endgroup$
– Yves Daoust
Jan 21 at 13:46
add a comment |
$begingroup$
For a diagram of the sum of consecutive odd nos, use the diagram in your sum of squares that shows a red L shape with an inner yellow L shape, with an inner blue L shape with an inner green L shape with a single green square.
answered Jan 21 at 11:13
jsc42jsc42
1
$endgroup$
$begingroup$
Can you please elaborate?
$endgroup$
– Michael Wang
Jan 22 at 0:29
add a comment |
$begingroup$
For a diagram of the sum of consecutive odd nos, use the diagram in your sum of squares that shows a red L shape with an inner yellow L shape, with an inner blue L shape with an inner green L shape with a single green square.
answered Jan 21 at 11:13
jsc42jsc42
1
$endgroup$
$begingroup$
Can you please elaborate?
$endgroup$
– Michael Wang
Jan 22 at 0:29
add a comment |
$begingroup$
For a diagram of the sum of consecutive odd nos, use the diagram in your sum of squares that shows a red L shape with an inner yellow L shape, with an inner blue L shape with an inner green L shape with a single green square.
answered Jan 21 at 11:13
jsc42jsc42
1
$endgroup$
For a diagram of the sum of consecutive odd nos, use the diagram in your sum of squares that shows a red L shape with an inner yellow L shape, with an inner blue L shape with an inner green L shape with a single green square.
answered Jan 21 at 11:13
jsc42jsc42
1
answered Jan 21 at 11:13
jsc42jsc42
1
answered Jan 21 at 11:13
jsc42jsc42
1
answered Jan 21 at 11:13
jsc42jsc42
1
1
$begingroup$
Can you please elaborate?
$endgroup$
– Michael Wang
Jan 22 at 0:29
add a comment |
$begingroup$
Can you please elaborate?
$endgroup$
– Michael Wang
Jan 22 at 0:29
$begingroup$
Can you please elaborate?
$endgroup$
– Michael Wang
Jan 22 at 0:29
$begingroup$
Can you please elaborate?
$endgroup$
– Michael Wang
Jan 22 at 0:29
add a comment |
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$begingroup$
In the first image, the white squares represent the terms of the sum $1+2+3+cdotcdotcdot+n$. Each square can be assumed to have unit area. The first row has $1$ white square, the second $2$ and so on. The sum is then merely the area of the white portion of the image, which is half the area of the $ntimes(n+1)$ rectangle
$endgroup$
– Shubham Johri
Jan 21 at 9:00
1
$begingroup$
I believe the second PWoW finds inspiration in the fact that $n^2$ is the sum of the first $n$ odd natural numbers.$$1^2+2^2+3^2+cdotcdotcdot+n^2=1+(1+3)+(1+3+5)+cdotcdotcdot+(1+3+5+cdotcdotcdot+2n-1)\=ntimes1+(n-1)times3+cdotcdotcdot+2times(2n-3)+1times(2n-1)$$
$endgroup$
– Shubham Johri
Jan 21 at 9:06
1
$begingroup$
The question didn't originally ask for a proof of the sum of consecutive odd numbers. (In general, you shouldn't add to a question once you have gotten answers, but be that as it may ...). That the sum of consecutive odd numbers is a square is actually illustrated with the colorful blocks on the left of the second image: note that the bands of color are the consecutive odd numbers.
$endgroup$
– Blue
Jan 21 at 9:40
1
$begingroup$
@Blue thank you. i will refrain from doing that next time
$endgroup$
– user8358234
Jan 21 at 9:41
$begingroup$
@user8358234: Good to know. :) The comment was intended both as (friendly!) advice to you, and as notice to future readers who might wonder why the existing answers don't address the last issue.
$endgroup$
– Blue
Jan 21 at 9:46