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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!

```

```
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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