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/
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