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[tsh44]

Hello I am also having rouboe with the second part of this word problem.

A 3-by-3-by-3 cube has 27 little cubes but also many other cubes of various sizes. How many cubes does it have in all? Find the toal number of cubes in an n-by-n-by-n cube.

Part 1:

3 x3 x3 cubes- 1 (1^3)
2 x 2 x 2 cubes- 8 (2^3)
1 x 1 x 1 cubes - 27(3^3)

So there is a total of 36 cubes.

But for Part B i am having trouble coming up with an expression or formula for the sum of the cubes. I know it is 1^3, 2^3, 3^ 3 in this case but I just can't come up with a rule for it. Thanks for your help.

 

[soroban]

Hello, tsh44!

A 3-by-3-by-3 cube has 27 little cubes but also many other cubes of various sizes.
How many cubes does it have in all?
Find the total number of cubes in an n-by-n-by-n cube.

Part 1:

3 x 3 x 3 cubes: 1 (1^3)
2 x 2 x 2 cubes: 8 (2^3)
1 x 1 x 1 cubes : 27(3^3)

So there is a total of 36 cubes. . . . good work!

The second part requires some advanced thinking or a lot of algebra.


You could start a list and hope to find a pattern.

n = 1
1 x 1 x 1 cubes: 1
\(\displaystyle \;\;\)Total: 1

n = 2
2 x 2 x 2 cubes: 1
1 x 1 x 1 cubes: 8
\(\displaystyle \;\;\)Total: 9

n = 3
3 x 3 x 3 cubes: 1
2 x 2 x 2 cubes: 8
1 x 1 x 1 cubes: 27
\(\displaystyle \;\;\)Total: 36

n = 4
4 x 4 x 4 cubes: 1
3 x 3 x 3 cubes: 8
2 x 2 x 2 cubes: 27
1 x 1 x 1 cubes: 64
\(\displaystyle \;\;\)Total: 100

They seem to be squares . . . but not consecutive squares.

They are: \(\displaystyle \,1^2,\:3^2,\:6^2,\:10^2,\:...\)

What is the basis for: 1, 3, 6, 10, ... ?

These are "triangular numbers": 1, 1+2, 1+2+3, 1+2+3+4, ...

The \(\displaystyle n^{th}\) triangular number is: \(\displaystyle \,T_n\:=\:\frac{n(n\,+\,1)}{2}\)


So an n-by-n-by-n cube contains a total of: \(\displaystyle \,\frac{n^2(n\,+\,1)^2}{4}\) cubes.