If the base is 120×120
120
×
120
units, and the height is 120
120
units, with the summit directly above the center of the base, then each side has a slope of 2
2
: one unit towards the center, the side wall rises by two units.
This means that at height 5
5
above the base, each side is 5×12=2.5
5
×
1
2
=
2.5
inwards, i.e. 115×115
115
×
115
in area, because each of the four sides (north, south, east west) is inwards by that amount. In other words, the bottommost layer of 5×5×5
5
×
5
×
5
blocks covers 115×115
115
×
115
unit area of the base, and thus consists of 23×23
23
×
23
blocks.
The next layer of blocks is similarly smaller, covering a 110×110
110
×
110
area on top of the previous layer of blocks, i.e. one block less in both directions, consisting of 22×22
22
×
22
blocks. (You might note that the center of each block in the upper layer is at the corners of four blocks in the lower layer, although it does not affect the math here at all.)
This continues upwards, with the 23rd layer having just one block. Note that since 23×5=115
23
×
5
=
115
, the summit of the pyramid (a rectangular pyramid itself, of course, with base 5×5
5
×
5
on top of the topmost block, and height another 5
5
) is empty; it does not have enough room to contain even a single block.
If we count the number of blocks in each layer, we get
=23×23+22×22+21×21+…+1
N
=
23
×
23
+
22
×
22
+
21
×
21
+
…
+
1
That is not too hard to do by hand (or, better, calculator; or even with 2×2 Lego bricks), but mathematically we can also say that
=∑=1232
N
=
∑
i
=
1
23
i
2
As it happens, it is already known (in lists of known sums) that
∑=12=16(+1)(2+1)
∑
i
=
1
n
i
2
=
1
6
n
(
n
+
1
)
(
2
n
+
1
)
and therefore
=16×23×24×47=4324