Derive a formula for the number of small square base pyramids required to create a bigger pyramid?

Question

To quote from the problem statement:

"Pyramids are built using smallest pyramids of "level 1", that are used as building blocks for higher levels. Stacking pyramids of "level 1" to create higher-level structures, leave gaps. Special parts, called "Tringus", are used to cover these gaps. They are of triangular shape to perfectly fill the gaps on pyramid sides."

Here is an image to help:

Explanation for this : "a pyramid of "Level 2" is constructed as follows: We are using a total of 6 pyramids of "level 1" (see the picture above). That leaves 4 gaps at the sides and we need 4 "Tringus" to fill them. In total 10 pieces."

Now I do understand the number of parts required for n^th level.

For Level 3, according to me, we have 9 (3*3) Level 1 triangles at the base.Then 4(2*2) inverted trianlges on the base and then just keep the "Level 2" triangle on it. That gives = 9 + 4 + 6(number of level 1 triangles used in level 2 as shown in the image) = 19. And the number of tringus required will be 3 on each face so 3*4=12. So total 19+12 = 31.

What I am unable to do is derive a general formula for number of total parts required for the k^th level. Any help on how to proceed?

Answer 1

I would compute the volume of a tringus in units of a pyramid to count the number of tringuses in each layer. The size $2$ pyramid has a volume of $8$. There are $6$ size $1$ pyramids in a size $2$ pyramid, so a tringus has a volume of $1/2$. Now we count the pyramids in a size $n$ pyramid
Upright pyramids-at level $k$ you have $k^2$ of them
Inverted pyramids-at level $k$ you have $(k-1)^2$ of them
Total pyramids $= \sum_{k=1}^n(k^2+(k-1)^2)=\frac 16(n(n+1)(2n+1)+(n-1)n(2n-1))=\frac 13(2n^3+n)$
The total volume is $n^3$, so the volume of tringuses is $\frac 13(n^3-n)$ and the number is $\frac 23(n^3-n)$.
The total number of pieces is $\frac 13(4n^3-n)$

Added: If you want to leave holes in the interior, just filling the openings on the faces, you will have $\frac12n(n−1)$ tringuses on each face, so $2n(n−1)$ total and the number of total pieces becomes $\frac13(2n3+n)+2n(n−1)$ This gives $31$ for $n=3$