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 48 mins ago
Blue
47.9k870152
asked 1 hour ago
user8358234user8358234
1568
$endgroup$
add a 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 48 mins ago
Blue
47.9k870152
asked 1 hour ago
user8358234user8358234
1568
$endgroup$
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
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
43 mins ago
1
$begingroup$
@Blue thank you. i will refrain from doing that next time
$endgroup$
– user8358234
42 mins ago
$begingroup$
@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
37 mins ago
add a 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 48 mins ago
Blue
47.9k870152
asked 1 hour ago
user8358234user8358234
1568
$endgroup$
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 48 mins ago
Blue
47.9k870152
asked 1 hour ago
user8358234user8358234
1568
edited 48 mins ago
Blue
47.9k870152
asked 1 hour ago
user8358234user8358234
1568
edited 48 mins ago
Blue
47.9k870152
edited 48 mins ago
Blue
47.9k870152
edited 48 mins ago
Blue
47.9k870152
47.9k870152
asked 1 hour ago
user8358234user8358234
1568
asked 1 hour ago
user8358234user8358234
1568
asked 1 hour ago
user8358234user8358234
1568
1568
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
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
43 mins ago
1
$begingroup$
@Blue thank you. i will refrain from doing that next time
$endgroup$
– user8358234
42 mins ago
$begingroup$
@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
37 mins ago
add a 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
1 hour ago
$begingroup$
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
1 hour ago
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
43 mins ago
1
$begingroup$
@Blue thank you. i will refrain from doing that next time
$endgroup$
– user8358234
42 mins ago
$begingroup$
@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
37 mins ago
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
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
43 mins ago
$begingroup$
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
43 mins ago
1
1
$begingroup$
@Blue thank you. i will refrain from doing that next time
$endgroup$
– user8358234
42 mins ago
$begingroup$
@Blue thank you. i will refrain from doing that next time
$endgroup$
– user8358234
42 mins ago
$begingroup$
@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
37 mins ago
$begingroup$
@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
37 mins ago
add a comment |
2 Answers
2
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 1 hour ago
Robert ZRobert Z
95k1063134
$endgroup$
$begingroup$
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
50 mins ago
$begingroup$
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
45 mins ago
$begingroup$
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
43 mins ago
$begingroup$
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
40 mins ago
$begingroup$
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
32 mins ago
|
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
$$4,(1+8+cdots n^3)=(n(n+1))^2.$$
answered 58 mins ago
Yves DaoustYves Daoust
125k671222
$endgroup$
$begingroup$
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
41 mins ago
$begingroup$
@user8358234: observe how the cells are displaced and see that their number is preserved.
$endgroup$
– Yves Daoust
39 mins ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
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 1 hour ago
Robert ZRobert Z
95k1063134
$endgroup$
$begingroup$
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
50 mins ago
$begingroup$
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
45 mins ago
$begingroup$
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
43 mins ago
$begingroup$
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
40 mins ago
$begingroup$
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
32 mins ago
|
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 1 hour ago
Robert ZRobert Z
95k1063134
$endgroup$
$begingroup$
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
50 mins ago
$begingroup$
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
45 mins ago
$begingroup$
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
43 mins ago
$begingroup$
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
40 mins ago
$begingroup$
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
32 mins ago
|
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 1 hour ago
Robert ZRobert Z
95k1063134
$endgroup$
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 1 hour ago
Robert ZRobert Z
95k1063134
edited 47 mins ago
answered 1 hour ago
Robert ZRobert Z
95k1063134
answered 1 hour ago
Robert ZRobert Z
95k1063134
answered 1 hour ago
Robert ZRobert Z
95k1063134
95k1063134
$begingroup$
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
50 mins ago
$begingroup$
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
45 mins ago
$begingroup$
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
43 mins ago
$begingroup$
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
40 mins ago
$begingroup$
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
32 mins ago
|
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?
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– user8358234
50 mins ago
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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 $(1^2+2^2+3^2+dots +n^2)=frac{n(n+1)(2n+1)}{6}$.
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– Robert Z
45 mins ago
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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?
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– user8358234
43 mins ago
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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.
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– Robert Z
40 mins ago
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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
32 mins ago
$begingroup$
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
50 mins ago
$begingroup$
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
50 mins ago
$begingroup$
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
45 mins ago
$begingroup$
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
45 mins ago
$begingroup$
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
43 mins ago
$begingroup$
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
43 mins ago
$begingroup$
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
40 mins ago
$begingroup$
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
40 mins ago
$begingroup$
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
32 mins ago
$begingroup$
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
32 mins ago
|
show 2 more comments
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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
$$4,(1+8+cdots n^3)=(n(n+1))^2.$$
answered 58 mins ago
Yves DaoustYves Daoust
125k671222
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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
41 mins ago
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@user8358234: observe how the cells are displaced and see that their number is preserved.
$endgroup$
– Yves Daoust
39 mins ago
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
$$4,(1+8+cdots n^3)=(n(n+1))^2.$$
answered 58 mins ago
Yves DaoustYves Daoust
125k671222
$endgroup$
$begingroup$
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
41 mins ago
$begingroup$
@user8358234: observe how the cells are displaced and see that their number is preserved.
$endgroup$
– Yves Daoust
39 mins ago
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
$$4,(1+8+cdots n^3)=(n(n+1))^2.$$
answered 58 mins ago
Yves DaoustYves Daoust
125k671222
$endgroup$
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
$$4,(1+8+cdots n^3)=(n(n+1))^2.$$
answered 58 mins ago
Yves DaoustYves Daoust
125k671222
edited 47 mins ago
answered 58 mins ago
Yves DaoustYves Daoust
125k671222
answered 58 mins ago
Yves DaoustYves Daoust
125k671222
answered 58 mins ago
Yves DaoustYves Daoust
125k671222
125k671222
$begingroup$
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
41 mins ago
$begingroup$
@user8358234: observe how the cells are displaced and see that their number is preserved.
$endgroup$
– Yves Daoust
39 mins ago
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
41 mins ago
$begingroup$
@user8358234: observe how the cells are displaced and see that their number is preserved.
$endgroup$
– Yves Daoust
39 mins ago
$begingroup$
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
41 mins ago
$begingroup$
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
41 mins ago
$begingroup$
@user8358234: observe how the cells are displaced and see that their number is preserved.
$endgroup$
– Yves Daoust
39 mins ago
$begingroup$
@user8358234: observe how the cells are displaced and see that their number is preserved.
$endgroup$
– Yves Daoust
39 mins ago
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
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– Shubham Johri
1 hour ago
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I believe the second PWoW finds inspiration in the fact that $n^2$ is the sum of the first $n$ odd natural numbers.$$1^2+2^2+3^2+cdotcdotcdot+n^2=1+(1+3)+(1+3+5)+cdotcdotcdot+(1+3+5+cdotcdotcdot+2n-1)\=ntimes1+(n-1)times3+cdotcdotcdot+2times(2n-3)+1times(2n-1)$$
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– Shubham Johri
1 hour ago
1
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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.
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– Blue
43 mins ago
1
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@Blue thank you. i will refrain from doing that next time
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– user8358234
42 mins ago
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@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.
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– Blue
37 mins ago