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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


2B 2B

3B 3B 3B

4B 4B 4B 4B

4B 4B 4B 4B

3B 3B 3B

  2B 2B


Then combine so

1B 4B 4B 4B 4B

2B 2B 3B 3B 3B

3B 3B 3B 2B 2B

4B 4B 4B 4B 1B

  • $\begingroup$ It means $6$ of each block type. $\endgroup$ Jan 15 '19 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 '19 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. $\endgroup$ Jan 15 '19 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 '19 at 16:14
  • $\begingroup$ I understand, but it is harder to visualize $3d$ blocks that way. $\endgroup$ Jan 15 '19 at 16:14

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/

  • $\begingroup$ You can accept this answer by clicking the tick mark next to it. $\endgroup$ Jan 15 '19 at 17:41
  • $\begingroup$ every time i do, it says i cannot accept my own answer $\endgroup$
    – user477465
    Jan 15 '19 at 17:43
  • $\begingroup$ it says i have to wait 2 days lol $\endgroup$
    – user477465
    Jan 15 '19 at 17:45
  • $\begingroup$ See Can I answer my own question? in the Help Center. $\endgroup$
    – NickD
    Jan 15 '19 at 22:05
  • $\begingroup$ yes but i have to wait 2 days before i can do it lol $\endgroup$
    – user477465
    Jan 15 '19 at 22:26

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