stbenge napsal(a):
> 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
you say:
 >  n*(n+1)*(n+2)*(n+3)/4
 >  -n*(n+1)*(n+2)/3
 >  -n*(n+1)/2
I guess:
 >  n*(n+1)*(n+2)*(n+3)/4
 >  -n*(n+1)*(n+2)/3
 >  +n*(n+1)/2
or:
 >  n*(n+1)*(n+2)*(n+3)/4
 >  -n*(n+1)*(n+2)/3

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