The sixth 3D cross shape has 61 cubes in its middle layer which is based on its corresponding 2D cross shape.
61=25+36
=5*5+6*6
By tabulating all the above result I can derive the general result.
The first column values of the table were always 1 less than the corresponding cross shape where as the second column values are the cross shape numbers broken up.
Therefore the general formula for the Nth cross shape must have (n-1)2+n2 cubes.
Mn=(n-1)2+n2 cubes in it
Mn=n2-2n+1+n2
Mn=2n2-2n+1, where
n cross shape level
Mn Number of cubes in the middle layer of nth cross-shape
However the result obtained earlier for 2D cross shapes and proved alternatively using triangular numbers but now I have realized that the 3D cross shapes are built up based on 2D cross shapes.
Since I was just able to derive the general result for the number of cubes in the I assumed that I will be able to use the same approach for all 3D cross shapes more over I have noticed that the number of cubes is the sum of the squares of any 2 successful natural numbers. Therefore I will try to take a similar approach towards the whole 3D shapes al well. I have already found out the number of cubes in any 3D cross shape now I am going to break up these values in such a way that they can be expressed as sum of squares of natural numbers in a sequential order. The numbers of cubes of 3D cross shape are given below:
1=1
7=1+5+1
25=1+5+13+5+1
63=1+5+13+25+13+5+1
129=1+5+13+25+41+25+13+5+1
231=1+5+13+25+41+61+41+25+13+5+1
It can be seen that the middle layer can be the plain of symmetry of any 3D cross shapes. There fore there will be equal number of cubes on equal sides of the middle layer.
As I can see from the above table the last value, eg. 11, 33… in each row the rest of the terms have formed a pattern these numbers basically are the sum of the squares of the natural numbers up to the corresponding cross shape value that is the sixth cross shape will have squares added from 1 to 6 therefore the nth cross shape will have squares added from 1 to n therefore the nth cross shape will have 12+22+32+42+….+n2+constant.
It is apparent that in order to rationalize the formula I have derived earlier for 3D cross shape. I must first find the common sum of squares of natural numbers.
My GCSE curriculum does not contain any method of finding this sum of squares which is why I have looked elsewhere to prove this result. I have taken from the book “Further mathematics 1” by Geoff Mannall and Michael Kenwood.
12+22+32+…...+n2=nΣr2
r=1
Where Σ (sigma) is the summation of terms using the identity 24r2+2Ξ (2r+1)3-(2r-1)3 and take r=1, 2, 3,…..n
When r=1; 24(1)2+2=33-13
r=2; 24(2)2+2=53-33
r=3; 24(3)2+2=73-53
r=n-1; 24(n-1)2+2= (2n-1)3-(2n-3)3
r=n; 24n2+2= (2n+1)3-(2n-3)3
By adding all the above right terms, the result is (2n+1)3-13 which is equal to
8n3+12n2+6n