alphaQuad wrote:
>> Yeah well, like I said, it was fun to come up with this independently. I
>> wonder if anyone else has used a similar geometric approach?
>>
>> Sam
> 
> you cant just leave it like that.
> 
> 4D pyramid = 1 + 2^3 + 3^3 + .. + n^3

> 
> 


Where did you get that long decimal number?

I am more of a hacker than a mathematician or even a programmer. Before 
that, I use art as a visualization tool. Somehow though, I think there 
must be an approach to finding the brick count in 2d and 3d pyramids 
without the use of long decimal numbers. I'm thinking something like 
this might work:

4d pyramid? =
  n*(n+1)*(n+2)*(n+3)/4
  -n*(n+1)*(n+2)/3
  -n*(n+1)/2

It would be an extension of my 3d method:

3d pyramid =
  n*(n+1)*(n+2)/2
  -n*(n+1)/2

Of course, that 4d pyramid expression is probably wrong. It seems too 
simple a solution. Any time something is too easy, I quickly find out 
that there was a reason...

Sam

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