Count the number of small cubes that are cut by having a regular hexagonal cross-section of a large cube. I'm having trouble with the following question:
A large cube is made up of 15625 small cubes. If the large cube is cut by as a plane so that the cross-section is a regular hexagon, how many small cubes are cut?
I also looked it up on google but nothing came up. I don't know if my approach is correct or not, but I thought I could find the number of un-cut small cubes in the question if I could find the pattern to the number of un-cut small cubes in cubes made up of 8, 27, 64 (and so on) small cubes. So I drew cubes like such and tried to count the unaffected small cubes, but it was really difficult when it got to a cube made up of 27 small cubes.
Sorry for my wording as English is not my native language. Could anyone give me a hint to solve this? Thanks!

 A: A short simple solution was provided by  @JaapScherphuis right at the start, in the comments.  However it seems to have been overlooked, so I will elaborate his answer:
We cut the large cube into $(2k+1)^3$ smaller cubes in the obvious way (so $k=12$ is the original case posed).  The layers of the cube are evenly spaced so they slice the cross sectional hexagon evenly into $n=2k+1$ slices parallel to each edge of the regular hexagon.
Note that as $2k+1$ is odd, there are parallel cuts either side of the main diagonals.  Thus if you mark out where parallel cuts start along an edge of the hexagon, the separations along the edge are equal, except for the last which is half the usual separation to the adjacent vertex. This means that as you move along the base edge, the diagonals alternate between going up left, and up right.
$k=1:$

The number of cubes sliced is simply the number of regions this hexagon is divided into.
Label the layers of the hexagon from the bottom to middle (but one) $r_0,r_1,\cdots,r_{k-1}$.  The bottom layer $r_0$ has $k+1$ hexagons, $k$ triangles below and $k+2$ triangles above.  In general $r_i$ has $k+1+i$ hexagons, $k+i$ triangles below and $k+i+2$ triangles above.  The middle layer has $2k+1$ hexagons, $2k$ triangles below and $2k$ triangles above.  Counting the top half as well, we get the number of cubes sliced as:
$$
6k+1+2\sum_{i=0}^{k-1} 3(k+i+1)=9k^2+9k+1=\frac{9n^2-5}{4}.
$$
