In order to see a pattern between the totals of the shapes, we can arrange information in a table.
As you can see from the above table we can come to a formula of:
T = 6n + 44
T is the stair total, n the stair number and 44 is the remaining number.
T = n + (n+1) + (n+2) + (n+10) + (n+11) + (n+20)
T = 6n + 44
I have tested the formula above on other stairs on a 10 x 10 grid but not on other grid sizes.
e.g 1+2+3+11+12+21 = 50
T = 6n + 44
T = 6x1 + 44
T = 50
On a 9x9 grid, instead of each square increasing by ten vertically, the numbers increase by 9. We can see that this reflects the size of the grid.
Using the same algebraic method as previously, we can work out the increase or decrease in the stair total when the grid is changed, with g being the grid number. (I.e. 9x9)
T = n + (n+1) + (n+2) + (n+g) + (n+g+1) + (n+2g)
T = 6n + 4g + 4.
The formula T = 6n + 4g + 4 can be tested on any grid.
E.G. – A 9x9 grid
1 + 2 + 3 + 10 + 11 + 19 = 46
T= 6n + 4g + 4
T = 6x1 + 4x9 + 4
T = 6 + 36 + 4
T = 46
I can also use the formula for the total of a 3-stair shape on any grid to predict the result of a 3-stair shape on another grid.
I predict that the total of a 3 stair shape with a stair number, n, of 14 on an 11x11 grid.
T = 6n + 4g + 4
T = 6x14 + 4x11 + 4
T = 84 + 44 + 4
T = 132
14 + 15 + 16 + 25 + 26 + 36 = 132
PART 2
Translations of vectors (a/0), (b/0)
The further right you go across the grid, the higher the total will be, and vice versa for when you go left.
Vector (a/0) horizontal by a units.
Each square has increased by 5, which is the number of spaces moved horizontally.
We can arrange a formula which, when we insert into the original stair, will work out the total of the new stair.
When doing the formula, we can simply add on to the formula for working out the total of a 3-step stair on any grid.
T = (n+a) + (n+1+a) +(n+2+a) +(n+g+a) + (n+g+1+a) + (n+2g+a)
T = 6n + 6a + 4g + 4
T equals 6 times the stair number, plus 6 times the number of units moved horizontally, plus 4 times the grid number, in this instance it is 10, plus 4.
T = 6x1 + 6x5 + 4x10 + 4
T = 6 + 30 + 40 + 4
T = 80
T = 6+7+8+16+17+26
T= 80
Vector (b/0) vertical by b units
Each square has increased by 30, which is the number of places that the stair number has moved up by, times the grid size, which is 10.
We can arrange a formula which, when we insert into the original stair, will work out the total of the new stair.
When doing the formula, we can simply add on to the formula for working out the total of a 3-step stair on any grid.
T = (n+3b) + (n+1+3b) + (n+2+3b) + (n+g+1+3b) + (n+2gt+3b) + (n+g+3b)
T = 6n + 18b + 4g + 4
T equals 6 times the stair number, plus 18 times the number of units moved vertically, plus 4 times the grid number, in this instance it is 10, plus 4.
T = 6x1 + 18x10 + 4x10 + 4
T = 6 + 180 + 40 + 4
T = 230
T = 31+41+51+32+33+42
T = 230
