4 x N = the number of white squares
4N = W
And I also worked out the equation for the Diameter of any pattern:
2 x N + 1 = the diameter of the pattern
2N+1 = D
These two equations were relatively easy to work out because they follow a set increment. I simply looked down the table above and noted down the difference in them:
Forming an equation for the total number of squares is slightly harder. Now I will explain how I worked it out:
(for the example I have used N equal to 4)
-
I drew more patterns of crosses
- I examined the structure of the crosses and fitted in the value of N with strips of squares. I did this N times to see how many squares would be left.
- I noticed that the number of squares left over after fitting in the value of N was actually a factor of the value of N so I divided the remaining squares by the value of N
- I noticed the single centre square so I need to add ‘+ 1’ to the equation formed
After considering all of these points I produced the following equation:
N2 + 6N + 1
In the example this would translate to:
(4 x 4) + (6 x 4) + 1 = Total Squares
16 + 24 + 1 = T
31 = T
There are 31 squares in this pattern.
Now I am going to test this formula with different sized patterns:
N=5
-
6N = 30 ✗This doesn’t work
In this pattern, there are 35 remaining squares after I have squared the value of N instead of 30 squares (6N). If I divide 35 by the value of N (5), the answer is 7. I predict that the coefficient of N is in some way related to
the value of N.
Now I will investigate this:
For N=4, the value of the coefficient was 6
For N=5, the value of the coefficient was 7
I predict that it follows the pattern of N + 2 and so therefore I predict that:
For N=6, the value of the coefficient will be 8
I can test this without having to draw a pattern by adding it to the equation:
N2 + 8N + 1 N=6
(6 x 6) + (8 x 6) + 1 = 36 + 48 + 1 = 85 ✓
Another Prediction:
N2 + 10N + 1 N=8
(8 x 8) + (10 x 8) + 1 = 64 + 80 + 1 = 145 ✓ Correct
Now I will add this to my equation:
N2 + 6N + 1
(Incorrect Version)
I need to change to coefficient to reflect ‘N + 2’
N2 + N(N+2) + 1
This is now correct but can be simplified:
N2 + N2 + 2N +1
And further:
2N2 + 2N + 1
This is now my final formula. I am going to test it on the same patterns I did before just in case I have made an error.
N=6 2N2 + 2N + 1 = 2(6 x 6) + (2 x 6) + 1
= (2 x 36) + 12 + 1
= 72 + 12 + 1
= 85 ✓ Correct
N=8 2N2 + 2N + 1 = 2(8 x 8) + (2 x 8) + 1
= (2 x 64) + 16 + 1
= 128 + 16 +1
= 145 ✓ Correct
Now I know this formula works, I am able to predict the number of squares in much bigger patterns:
N=20 2N2 + 2N + 1 = 2(20 X 20) + (2 X 20) + 1
= (2 X 400) + 40 + 1
= 800 + 40 + 1
= 841 total squares