Cubes in a Big Cube

Date: 03/11/2002 at 09:45:48
From: Yannik
Subject: Cubes in a big cube

Dr. Math, 

I know the formula to find out the number of squares in an n*n square, 
which is n(n+1)(2n+1)/6. Is there a formula for the number of cubes in 
an n*n*n cube?  

Thanks.

Date: 03/11/2002 at 12:17:37
From: Doctor Peterson
Subject: Re: Cubes in a big cube

Hi, Yannik.

You probably know that the formula you gave is the formula for the sum 
of consecutive squares:

    1 + 4 + 9 + ... + n^2 = n(n+1)(2n+1)/6

This is the number of squares in an n*n square, because it contains 
1 n*n square, 2^2 (n-1)*(n-1) squares, and so on down to n^2 1*1 
squares.

The same is true for a cube; you want to find

    1 + 8 + 27 + ... + n^3

You can find this by searching our archives for "sum cubes":

   Triangular Numbers in a Proof
   http://mathforum.org/dr.math/problems/samflan.4.8.97.html   

It is

    1 + 8 + 27 + ... + n^3 = (1 + 2 + 3 + ... + n)^2

                           = [n(n+1)/2]^2

For example, a 3*3 cube has 1 3*3, 8 2*2, and 27 1*1 cubes, making a 
total of 36, which is the same as [3*4/2]^2 = 6^2.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/