PLEASE HELP, no one will answer, please

a figure composed of congruent cubic boxes has four layers. On the lowest layer are 7 rows of 7 blocks each. Centered on the bottom layer or 5 rows of 5 blocks each. Centered on top of that are 3 rows of 3 blocks each. Finally, a central block is placed in the entire structure. Then the figure is painted, except for the bottom. How many blocks have 6, 5, 4, 3, 2, 1, and 0 painted faces? What's the rule for finding out how many layers there are, supposing a figure has n rows and n blocks, each on the bottom layer? What's the rule for finding out how many blocks have 3, 2, and 0 painted faces?

thanks so much!! Please answer thoroughly, no mean comments like do it yourself please!

Measure out and draw a square 7 units by 7.
Fit inside that the 5 by 5 which is just 1 unit in all the way round.
and now the 3 by 3 square which fits in that similarly,
and finally a central block 1 by 1
You now have a birds-eye view of your stack of boxes.

Imagine yourself painting the boxes but each day you just do
one lot of outside edges or one lot of level surfaces.

1) Outside edges of base layer. 7*4 = 28 faces

2) Level surfaces top of base layer. 7 + 5 + 7 + 5 = 24 faces

3) Outside edges of second layer. 5*4 = 20 faces

4) Level surfaces top of second layer. 5 + 3 + 5 + 3 = 16 faces

5) Outside edges of third layer. 3*4 = 12 faces

6) Level surfaces top of third layer 3 + 1 + 3 + 1 = 8 faces

7) Outside edges of top layer 1*4 = 4 faces

8) Top of top layer = 1 face

Now it is possible to see a pattern and then work out a formula
All except the last one form a shrinking arithmetic progression
Total = (28 + 24 + 20 + 16 + 12 + 8 + 4) + 1 = 113

If you like formulae, AP = (last number + first number)/2 * number of numbers

AP = (28 + 4)/2 * 7 = 112 and then add on the one for the top to make 113

Regards - Ian