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Cubes in a Grid


Date: 06/17/99 at 15:28:17
From: Daniel Holman
Subject: Grids

I'm trying to find a formula to determine how many cubes you need for 
a certain size grid. The bottom row in the cube always has 1 cube, the 
second has 5, etc. I have to work out how many cubes you need in all 
for a certain size grid. For example, a 3x3 grid needs 1 on the 
bottom, 5 in the middle and another 1 on the top. So that's 7 cubes 
for a 3x3 grid.  I will give you the figures for larger ones: 
3x3 = 7,  5x5 = 25, 7x7 = 63, 9x9 = 129, 11x11 = 231. I just cannot 
work out a formula to determine how many cubes you need for a specific 
size grid. Arrgghh! 

Please, please help!


Date: 06/18/99 at 10:51:52
From: Doctor Anthony
Subject: Re: Grids

Make up a difference table

       n =    1       2       3        4        5        6
   size  =   1x1     3x3     5x5      7x7      9x9     11x11
    f(n) =    1       7      25        63      129      231

1st diff.         6       18       38       66      102
2nd diff.             12       20       28      36
3rd diff.                  8        8        8

With 3rd differences constant the expression for the nth term will be 
a cubic polynomial in n.

So we assume  f(n) = an^3 + bn^2 + cn + d

     n = 1   gives     a  +  b  +  c  +  d  = 1
     n = 2     "      8a  + 4b  + 2c  +  d  = 7
     n = 3     "     27a  + 9b  + 3c  +  d  = 25
     n = 4     "     64a  +16b  + 4c  +  d  = 63 

So we have four equations with four unknowns. The solutions are

   a=4/3    b=-2     c=8/3     d=-1

Therefore  f(n) = 4n^3/3 - 2n^2 + 8n/3 - 1

                  4n^3 - 6n^2 + 8n - 3
           f(n) = --------------------
                          3

Check this answer for n = 6

                   864 - 216 + 48 - 3      
           f(6) = --------------------  =  231
                           3

which checks with the value in the table.

We could also express the formula in terms of grid size by replacing n 
in the above formula by (s+1)/2

               (1/2)(s+1)^3 - (3/2)(s+1)^2 + 4(s+1) - 3
     so f(s) = ----------------------------------------
                             3

                 (s+1)^3 - 3(s+1)^2 + 8(s+1) - 6
        f(s) =  ---------------------------------
                              6

                 s^3 + 5s 
        f(s) = -----------   =  s(s^2 + 5)/6
                    6

check this result for an 11x11 grid, that is s=11
                
     f(s) =  (1/6) x 11 x (121+5) =  11 x 126/6 =  231   which checks
                          
- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Sequences, Series

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