The previous shape is the extended version of the original cross shape it is surrounded by 16 outer layer squares and contains a total of 41 squares. I will extend it one more time, which will look like this:
The shape above is surrounded by 20 blank outer layer squares and 41 inner dark squares making a total number of 61 squares.
The basic origin of the cross shape until the 6th level.
Level 1
Level 2
Level 3
Level 4
Level 5
Level 6
The numbers of squares added are the multiples of four and occur in this order:
4 * 0 = 0
4 * 1 = 4
4 * 2 = 8
4 * 3 = 12
4 * 4 = 16
4 * 5 = 20
Therefore I can see a sequence in the order of these differences and the common difference between them is four.
Therefore the difference between the number of squares in the 6th and the 7th level will be 24 and the difference between the 7th and 8th will be 28.
Therefore 61 + 24 = 85
The 7th level contains 85 squares
85 + 28 = 113
Hence, the 8th level will contains 113 squares
Level 7
Level 8
The 8th level comprises of 113 squares, this means that there are 28 more squares.
Here I will prove my forecasts by extracting the same conclusion in diagrammatical way;
Level 4
Level 5
The line in the fourth level crosses 3 squares on the outer layer of the figure. The lines all cross through the blank squares only. In the diagram above, the four edge squares are crossed twice by the lines.
The line in the second diagram crosses 4 squares on the outer layer of the figure. These squares (4 edge squares) are counted twice because two lines pass through them.
In Level 4 each line crosses through 4 squares similarly, in the fifth level crosses five squares there I can say that the number of squares crossed by each line is equal to the level’s number.
Since the four outer layer squares are counted twice these squares must be subtracted.
Therefore I can construct the following formulae: Un =4n – 4
UN is the number of squares added to make the same level, where n is the number of the level. The four that is multiplied by n is the number of lines in each diagram.
Trial in Level 4
4n – 4 = Number of added squares
4 x 4 – 4 = number of squares added
16 – 4 = 12
12 squares were added to make level four
Level 1 2 3 4 5 6
Total squares: 1 5 13 25 41 61
Difference 4 8 12 16 20
2nd difference 4 4 4 4
When the 1st difference is subtracted from the previous value the result will always be four. If the difference is found in the second line it is a quadratic equation which is in the form of:
UN=an2+bn+c
From the above formula above we can obtain an equation which would decide the number of squares in any level of the cross shape.
‘a’ represents half of the constant difference
‘Un’ is the number of squares in any term,
n = the number of the level of the diagram
b and c need to be found in the equation to solve the equation.
Constant difference = 4
Therefore ‘a’ = ½ x 4
a = 2
When n = 1
If n=1, 1= 2(1)2 b+(1)+c
1= 2+b+c
b+c = 1-2= -1
if n=2, 5=2(2)2 + b(2)+c
5=8+2b+c
2b+c= 5-8= -3
Solve simultaneously= equation 2 -1
= 2b+c= -3 1
-b+c= -1 2
b=-2
2b+c= -3
2(-2) +c= -3
-4 + c= -3
C= -3+4= 1
a = 2
b= -2
c = 1
Consequently, Un = an2 +bn +c
UN= 2n2 -2n +1
Verify in level 4
Un = 2n2 + -2n + 1
2(4)2 + -2(4) + 1 = 25
2 x 16 – 8 + 1 = 25
32 - 8 + 1 = 25
32–7= 25
25 = 25
The previous calculation proves that the formula is accurate. Now I will prove it diagrammatically.
1
3
5
1 3
The first row has 1 square and therefore I have marked the row by an arrow.
Therefore I can get the formulae 4(n – 1) = the added squares.
Here I will further my endeavors to the anlysation and investigation to Three Dimensional shapes.
