Mixing
How can I use the sum of squares formula to create blocks of a certain dimension?
$begingroup$
The previous expression was just the sum of consecutive squares so $1^2+2^2+...+n^2 = frac{n(n+1)(2n+1)}{6}$
I know how to derive this formula but can someone please explain the claim that
"This expression says that a box with dimensions $n(n+1)(2n+1)$ should contain six copies of $1^2+2^2+...+n^2$"
For a box that has a 4 x 5 x 9 dimension, $n=4$
So I have to construct a video out of individual $1^2+2^2+3^2+4^2$ building units
I have no idea what this means. $1^2+2^2+3^2+4^2 = 30$ cubes in total and 4 x 5 x 9 = 180 unit squares.
So am I supposed to build this with $1^2$ unit squares and $2^2 = 4$, 2 x 2 squares and $3^2=9$, 3 x 3 squares and $4^2 = 16$ 4 x 4 squares? and if so how can I organize it so that the box has dimensions 4 x 5 x 9?
I don't know how to organize the blocks and I don't know if my interpretation of the problem is correct?
EDIT: He also says that there should be 6 units? That means only six blocks? How is he getting that?
EDIT: stacking blocks
1B
2B 2B
3B 3B 3B
4B 4B 4B 4B
4B 4B 4B 4B
3B 3B 3B
2B 2B
1B
Then combine so
1B 4B 4B 4B 4B
2B 2B 3B 3B 3B
3B 3B 3B 2B 2B
4B 4B 4B 4B 1B
algebra-precalculus
asked Jan 15 at 15:40
user477465user477465
200111
$endgroup$
|
show 9 more comments
$begingroup$
The previous expression was just the sum of consecutive squares so $1^2+2^2+...+n^2 = frac{n(n+1)(2n+1)}{6}$
I know how to derive this formula but can someone please explain the claim that
"This expression says that a box with dimensions $n(n+1)(2n+1)$ should contain six copies of $1^2+2^2+...+n^2$"
For a box that has a 4 x 5 x 9 dimension, $n=4$
So I have to construct a video out of individual $1^2+2^2+3^2+4^2$ building units
I have no idea what this means. $1^2+2^2+3^2+4^2 = 30$ cubes in total and 4 x 5 x 9 = 180 unit squares.
So am I supposed to build this with $1^2$ unit squares and $2^2 = 4$, 2 x 2 squares and $3^2=9$, 3 x 3 squares and $4^2 = 16$ 4 x 4 squares? and if so how can I organize it so that the box has dimensions 4 x 5 x 9?
I don't know how to organize the blocks and I don't know if my interpretation of the problem is correct?
EDIT: He also says that there should be 6 units? That means only six blocks? How is he getting that?
EDIT: stacking blocks
1B
2B 2B
3B 3B 3B
4B 4B 4B 4B
4B 4B 4B 4B
3B 3B 3B
2B 2B
1B
Then combine so
1B 4B 4B 4B 4B
2B 2B 3B 3B 3B
3B 3B 3B 2B 2B
4B 4B 4B 4B 1B
algebra-precalculus
asked Jan 15 at 15:40
user477465user477465
200111
$endgroup$
$begingroup$
It means $6$ of each block type.
$endgroup$
– Mohammad Zuhair Khan
Jan 15 at 15:51
$begingroup$
6 1 x 1, 6 2 x 2, 6 3 x 3 and 6 4 x 4?
$endgroup$
– user477465
Jan 15 at 16:01
$begingroup$
Yes, though it seems difficult enough for me to solve without having the blocks infront of me. Geogebra 3d fails me as well. Do you know any software that can do the trick? I am willing to place the blocks manually, but graphing each block is way too tedious.
$endgroup$
– Mohammad Zuhair Khan
Jan 15 at 16:04
$begingroup$
maybe just write out 1B, 2B, 3B, 4B for the blocks representing 1 x1, 2 x 2, etc. and then stack them up? I can show you an example if I'm unclear on what I'm saying.
$endgroup$
– user477465
Jan 15 at 16:14
$begingroup$
I understand, but it is harder to visualize $3d$ blocks that way.
$endgroup$
– Mohammad Zuhair Khan
Jan 15 at 16:14
|
show 9 more comments
$begingroup$
The previous expression was just the sum of consecutive squares so $1^2+2^2+...+n^2 = frac{n(n+1)(2n+1)}{6}$
I know how to derive this formula but can someone please explain the claim that
"This expression says that a box with dimensions $n(n+1)(2n+1)$ should contain six copies of $1^2+2^2+...+n^2$"
For a box that has a 4 x 5 x 9 dimension, $n=4$
So I have to construct a video out of individual $1^2+2^2+3^2+4^2$ building units
I have no idea what this means. $1^2+2^2+3^2+4^2 = 30$ cubes in total and 4 x 5 x 9 = 180 unit squares.
So am I supposed to build this with $1^2$ unit squares and $2^2 = 4$, 2 x 2 squares and $3^2=9$, 3 x 3 squares and $4^2 = 16$ 4 x 4 squares? and if so how can I organize it so that the box has dimensions 4 x 5 x 9?
I don't know how to organize the blocks and I don't know if my interpretation of the problem is correct?
EDIT: He also says that there should be 6 units? That means only six blocks? How is he getting that?
EDIT: stacking blocks
1B
2B 2B
3B 3B 3B
4B 4B 4B 4B
4B 4B 4B 4B
3B 3B 3B
2B 2B
1B
Then combine so
1B 4B 4B 4B 4B
2B 2B 3B 3B 3B
3B 3B 3B 2B 2B
4B 4B 4B 4B 1B
algebra-precalculus
asked Jan 15 at 15:40
user477465user477465
200111
$endgroup$
The previous expression was just the sum of consecutive squares so $1^2+2^2+...+n^2 = frac{n(n+1)(2n+1)}{6}$
I know how to derive this formula but can someone please explain the claim that
"This expression says that a box with dimensions $n(n+1)(2n+1)$ should contain six copies of $1^2+2^2+...+n^2$"
For a box that has a 4 x 5 x 9 dimension, $n=4$
So I have to construct a video out of individual $1^2+2^2+3^2+4^2$ building units
I have no idea what this means. $1^2+2^2+3^2+4^2 = 30$ cubes in total and 4 x 5 x 9 = 180 unit squares.
So am I supposed to build this with $1^2$ unit squares and $2^2 = 4$, 2 x 2 squares and $3^2=9$, 3 x 3 squares and $4^2 = 16$ 4 x 4 squares? and if so how can I organize it so that the box has dimensions 4 x 5 x 9?
I don't know how to organize the blocks and I don't know if my interpretation of the problem is correct?
EDIT: He also says that there should be 6 units? That means only six blocks? How is he getting that?
EDIT: stacking blocks
1B
2B 2B
3B 3B 3B
4B 4B 4B 4B
4B 4B 4B 4B
3B 3B 3B
2B 2B
1B
Then combine so
1B 4B 4B 4B 4B
2B 2B 3B 3B 3B
3B 3B 3B 2B 2B
4B 4B 4B 4B 1B
algebra-precalculus
algebra-precalculus
asked Jan 15 at 15:40
user477465user477465
200111
asked Jan 15 at 15:40
user477465user477465
200111
edited Jan 15 at 16:16
user477465
asked Jan 15 at 15:40
user477465user477465
200111
asked Jan 15 at 15:40
user477465user477465
200111
asked Jan 15 at 15:40
user477465user477465
200111
200111
$begingroup$
It means $6$ of each block type.
$endgroup$
– Mohammad Zuhair Khan
Jan 15 at 15:51
$begingroup$
6 1 x 1, 6 2 x 2, 6 3 x 3 and 6 4 x 4?
$endgroup$
– user477465
Jan 15 at 16:01
$begingroup$
Yes, though it seems difficult enough for me to solve without having the blocks infront of me. Geogebra 3d fails me as well. Do you know any software that can do the trick? I am willing to place the blocks manually, but graphing each block is way too tedious.
$endgroup$
– Mohammad Zuhair Khan
Jan 15 at 16:04
$begingroup$
maybe just write out 1B, 2B, 3B, 4B for the blocks representing 1 x1, 2 x 2, etc. and then stack them up? I can show you an example if I'm unclear on what I'm saying.
$endgroup$
– user477465
Jan 15 at 16:14
$begingroup$
I understand, but it is harder to visualize $3d$ blocks that way.
$endgroup$
– Mohammad Zuhair Khan
Jan 15 at 16:14
|
show 9 more comments
$begingroup$
It means $6$ of each block type.
$endgroup$
– Mohammad Zuhair Khan
Jan 15 at 15:51
$begingroup$
6 1 x 1, 6 2 x 2, 6 3 x 3 and 6 4 x 4?
$endgroup$
– user477465
Jan 15 at 16:01
$begingroup$
Yes, though it seems difficult enough for me to solve without having the blocks infront of me. Geogebra 3d fails me as well. Do you know any software that can do the trick? I am willing to place the blocks manually, but graphing each block is way too tedious.
$endgroup$
– Mohammad Zuhair Khan
Jan 15 at 16:04
$begingroup$
maybe just write out 1B, 2B, 3B, 4B for the blocks representing 1 x1, 2 x 2, etc. and then stack them up? I can show you an example if I'm unclear on what I'm saying.
$endgroup$
– user477465
Jan 15 at 16:14
$begingroup$
I understand, but it is harder to visualize $3d$ blocks that way.
$endgroup$
– Mohammad Zuhair Khan
Jan 15 at 16:14
$begingroup$
It means $6$ of each block type.
$endgroup$
– Mohammad Zuhair Khan
Jan 15 at 15:51
$begingroup$
It means $6$ of each block type.
$endgroup$
– Mohammad Zuhair Khan
Jan 15 at 15:51
$begingroup$
6 1 x 1, 6 2 x 2, 6 3 x 3 and 6 4 x 4?
$endgroup$
– user477465
Jan 15 at 16:01
$begingroup$
6 1 x 1, 6 2 x 2, 6 3 x 3 and 6 4 x 4?
$endgroup$
– user477465
Jan 15 at 16:01
$begingroup$
Yes, though it seems difficult enough for me to solve without having the blocks infront of me. Geogebra 3d fails me as well. Do you know any software that can do the trick? I am willing to place the blocks manually, but graphing each block is way too tedious.
$endgroup$
– Mohammad Zuhair Khan
Jan 15 at 16:04
$begingroup$
Yes, though it seems difficult enough for me to solve without having the blocks infront of me. Geogebra 3d fails me as well. Do you know any software that can do the trick? I am willing to place the blocks manually, but graphing each block is way too tedious.
$endgroup$
– Mohammad Zuhair Khan
Jan 15 at 16:04
$begingroup$
maybe just write out 1B, 2B, 3B, 4B for the blocks representing 1 x1, 2 x 2, etc. and then stack them up? I can show you an example if I'm unclear on what I'm saying.
$endgroup$
– user477465
Jan 15 at 16:14
$begingroup$
maybe just write out 1B, 2B, 3B, 4B for the blocks representing 1 x1, 2 x 2, etc. and then stack them up? I can show you an example if I'm unclear on what I'm saying.
$endgroup$
– user477465
Jan 15 at 16:14
$begingroup$
I understand, but it is harder to visualize $3d$ blocks that way.
$endgroup$
– Mohammad Zuhair Khan
Jan 15 at 16:14
$begingroup$
I understand, but it is harder to visualize $3d$ blocks that way.
$endgroup$
– Mohammad Zuhair Khan
Jan 15 at 16:14
|
show 9 more comments
1 Answer
1
active
oldest
votes
$begingroup$
I can make 6 pyramid like structures starting with a base of 4 x 4, then on top 3 x 3, then 2 x 2 then 1 x 1. then I can combine them all. I found this from
https://ckrao.wordpress.com/2012/03/14/the-sum-of-consecutive-squares-formula/
answered Jan 15 at 17:31
user477465user477465
200111
$endgroup$
$begingroup$
You can accept this answer by clicking the tick mark next to it.
$endgroup$
– Mohammad Zuhair Khan
Jan 15 at 17:41
$begingroup$
every time i do, it says i cannot accept my own answer
$endgroup$
– user477465
Jan 15 at 17:43
$begingroup$
it says i have to wait 2 days lol
$endgroup$
– user477465
Jan 15 at 17:45
$begingroup$
See Can I answer my own question? in the Help Center.
$endgroup$
– NickD
Jan 15 at 22:05
$begingroup$
yes but i have to wait 2 days before i can do it lol
$endgroup$
– user477465
Jan 15 at 22:26
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3074562%2fhow-can-i-use-the-sum-of-squares-formula-to-create-blocks-of-a-certain-dimension%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I can make 6 pyramid like structures starting with a base of 4 x 4, then on top 3 x 3, then 2 x 2 then 1 x 1. then I can combine them all. I found this from
https://ckrao.wordpress.com/2012/03/14/the-sum-of-consecutive-squares-formula/
answered Jan 15 at 17:31
user477465user477465
200111
$endgroup$
$begingroup$
You can accept this answer by clicking the tick mark next to it.
$endgroup$
– Mohammad Zuhair Khan
Jan 15 at 17:41
$begingroup$
every time i do, it says i cannot accept my own answer
$endgroup$
– user477465
Jan 15 at 17:43
$begingroup$
it says i have to wait 2 days lol
$endgroup$
– user477465
Jan 15 at 17:45
$begingroup$
See Can I answer my own question? in the Help Center.
$endgroup$
– NickD
Jan 15 at 22:05
$begingroup$
yes but i have to wait 2 days before i can do it lol
$endgroup$
– user477465
Jan 15 at 22:26
add a comment |
$begingroup$
I can make 6 pyramid like structures starting with a base of 4 x 4, then on top 3 x 3, then 2 x 2 then 1 x 1. then I can combine them all. I found this from
https://ckrao.wordpress.com/2012/03/14/the-sum-of-consecutive-squares-formula/
answered Jan 15 at 17:31
user477465user477465
200111
$endgroup$
$begingroup$
You can accept this answer by clicking the tick mark next to it.
$endgroup$
– Mohammad Zuhair Khan
Jan 15 at 17:41
$begingroup$
every time i do, it says i cannot accept my own answer
$endgroup$
– user477465
Jan 15 at 17:43
$begingroup$
it says i have to wait 2 days lol
$endgroup$
– user477465
Jan 15 at 17:45
$begingroup$
See Can I answer my own question? in the Help Center.
$endgroup$
– NickD
Jan 15 at 22:05
$begingroup$
yes but i have to wait 2 days before i can do it lol
$endgroup$
– user477465
Jan 15 at 22:26
add a comment |
$begingroup$
I can make 6 pyramid like structures starting with a base of 4 x 4, then on top 3 x 3, then 2 x 2 then 1 x 1. then I can combine them all. I found this from
https://ckrao.wordpress.com/2012/03/14/the-sum-of-consecutive-squares-formula/
answered Jan 15 at 17:31
user477465user477465
200111
$endgroup$
I can make 6 pyramid like structures starting with a base of 4 x 4, then on top 3 x 3, then 2 x 2 then 1 x 1. then I can combine them all. I found this from
https://ckrao.wordpress.com/2012/03/14/the-sum-of-consecutive-squares-formula/
answered Jan 15 at 17:31
user477465user477465
200111
answered Jan 15 at 17:31
user477465user477465
200111
answered Jan 15 at 17:31
user477465user477465
200111
answered Jan 15 at 17:31
user477465user477465
200111
200111
$begingroup$
You can accept this answer by clicking the tick mark next to it.
$endgroup$
– Mohammad Zuhair Khan
Jan 15 at 17:41
$begingroup$
every time i do, it says i cannot accept my own answer
$endgroup$
– user477465
Jan 15 at 17:43
$begingroup$
it says i have to wait 2 days lol
$endgroup$
– user477465
Jan 15 at 17:45
$begingroup$
See Can I answer my own question? in the Help Center.
$endgroup$
– NickD
Jan 15 at 22:05
$begingroup$
yes but i have to wait 2 days before i can do it lol
$endgroup$
– user477465
Jan 15 at 22:26
add a comment |
$begingroup$
You can accept this answer by clicking the tick mark next to it.
$endgroup$
– Mohammad Zuhair Khan
Jan 15 at 17:41
$begingroup$
every time i do, it says i cannot accept my own answer
$endgroup$
– user477465
Jan 15 at 17:43
$begingroup$
it says i have to wait 2 days lol
$endgroup$
– user477465
Jan 15 at 17:45
$begingroup$
See Can I answer my own question? in the Help Center.
$endgroup$
– NickD
Jan 15 at 22:05
$begingroup$
yes but i have to wait 2 days before i can do it lol
$endgroup$
– user477465
Jan 15 at 22:26
$begingroup$
You can accept this answer by clicking the tick mark next to it.
$endgroup$
– Mohammad Zuhair Khan
Jan 15 at 17:41
$begingroup$
You can accept this answer by clicking the tick mark next to it.
$endgroup$
– Mohammad Zuhair Khan
Jan 15 at 17:41
$begingroup$
every time i do, it says i cannot accept my own answer
$endgroup$
– user477465
Jan 15 at 17:43
$begingroup$
every time i do, it says i cannot accept my own answer
$endgroup$
– user477465
Jan 15 at 17:43
$begingroup$
it says i have to wait 2 days lol
$endgroup$
– user477465
Jan 15 at 17:45
$begingroup$
it says i have to wait 2 days lol
$endgroup$
– user477465
Jan 15 at 17:45
$begingroup$
See Can I answer my own question? in the Help Center.
$endgroup$
– NickD
Jan 15 at 22:05
$begingroup$
See Can I answer my own question? in the Help Center.
$endgroup$
– NickD
Jan 15 at 22:05
$begingroup$
yes but i have to wait 2 days before i can do it lol
$endgroup$
– user477465
Jan 15 at 22:26
$begingroup$
yes but i have to wait 2 days before i can do it lol
$endgroup$
– user477465
Jan 15 at 22:26
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3074562%2fhow-can-i-use-the-sum-of-squares-formula-to-create-blocks-of-a-certain-dimension%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
It means $6$ of each block type.
$endgroup$
– Mohammad Zuhair Khan
Jan 15 at 15:51
$begingroup$
6 1 x 1, 6 2 x 2, 6 3 x 3 and 6 4 x 4?
$endgroup$
– user477465
Jan 15 at 16:01
$begingroup$
Yes, though it seems difficult enough for me to solve without having the blocks infront of me. Geogebra 3d fails me as well. Do you know any software that can do the trick? I am willing to place the blocks manually, but graphing each block is way too tedious.
$endgroup$
– Mohammad Zuhair Khan
Jan 15 at 16:04
$begingroup$
maybe just write out 1B, 2B, 3B, 4B for the blocks representing 1 x1, 2 x 2, etc. and then stack them up? I can show you an example if I'm unclear on what I'm saying.
$endgroup$
– user477465
Jan 15 at 16:14
$begingroup$
I understand, but it is harder to visualize $3d$ blocks that way.
$endgroup$
– Mohammad Zuhair Khan
Jan 15 at 16:14