3 Step-Staircase / Grid Width 12
Suspected formula: T = 6n + 52
Prediction / Test: 6 x 20 + 52 = 172
20 + 21 + 22 + 32 + 33 + 44 = 172
Algebraic Proof:
n + (n+1) + (n+2) + (n+12) + (n+13) + (n+24) = 6n + 48
2 Step-Staircase
2-Step Staircase/ Grid Width 8
Suspected formula: T = 3n + 9
Prediction / Test: 3 x 20 + 9 = 69
20 + 21 + 28 = 69
Algebraic Proof:
n + (n+1) + (n+8) = 3n + 9
2-Step Staircase/ Grid Width 9
Suspected formula: T = 3n + 10
Prediction / Test: 3 x 20 + 10 = 70
20 + 21 + 29 = 70
Algebraic Proof:
n + (n+1) + (n+9) = 3n + 10
2-Step Staircase/ Grid Width 10
Suspected formula: T = 3n + 11
Prediction / Test: 3 x 31 + 11 = 104
31 + 32 + 41 = 104
Algebraic Proof:
n + (n+1) + (n+10) = 3n + 11
Here, I can sense a pattern: It increases by 1 on the number term every time the grid size increases so I predict that for 11, it will be: T = 3n + 12
2-Step Staircase/ Grid Width 11
Suspected formula: T = 3n + 12
Prediction / Test: 3 x 20 + 12 = 72
20 + 21 + 31 = 72
Algebraic Proof:
n + (n+1) + (n+11) = 3n + 12
2-Step Staircase/ Grid Width 12
Suspected formula: T = 3n + 13
Prediction / Test: 3 x 20 + 13 = 73
20 + 21 + 32 = 73
Algebraic Proof:
n + (n+1) + (n+12) = 3n + 13
1 Step-Staircase
For 1 Step-Staircases it will always be T = n because the n is the bottom left number and the formula is not accumulating anything but the bottom left number which is the only term in the formula.
4 Step-Staircase
4 Step-Staircase / Grid Width 8
Suspected formula: T = 10n + 90
Prediction / Test: 10 x 25 + 90 = 340
25 + 26 + 27 + 28 + 33 + 34 + 35 + 41 + 42 + 49 = 340
Algebraic Proof:
n + (n+1) + (n+2) + (n+3) + (n+8) + (n+9) + (n+10) + (n+16) + (n+17) + (n+24) = 10n + 90
4 Step-Staircase / Grid Width 9
Suspected formula: T = 10n + 100
Prediction / Test: 10x 24 + 100 = 340
24 + 25 + 26 + 27 + 33 + 34 + 35 + 42 + 43 + 51 = 340
Algebraic Proof:
n + (n+1) + (n+2) + (n+3) + (n+9) + (n+10) + (n+11) + (n+18) + (n+19) + (n+27) = 10n + 100
4 Step-Staircase / Grid Width 10
Suspected formula: T = 10n + 110
Prediction / Test: 10 x 23 + 110 = 340
23 + 24 + 25 + 26 + 33 + 34 + 35 + 43 + 44 + 53 = 340
Algebraic Proof:
n + (n+1) + (n+2) + (n+3) + (n+10) + (n+11) + (n+12) + (n+20) + (n+21) + (n+30) = 10n + 110
From here, I see a pattern of a jump size of 10 so I predict for grid widths 11 and 12 will be T = 10n + 120 and T = 10n + 130.
4 Step-Staircase / Grid Width 11
Suspected formula: T = 10n + 120
Prediction / Test: 10 x 47 + 120 = 590
47 + 48 + 49 + 50 + 58 + 59 + 60 + 69 + 70 + 80 = 590
Algebraic Proof:
n + (n+1) + (n+2) + (n+3) + (n+11) + (n+12) + (n+13) + (n+22) + (n+23) + (n+33) = 10n + 120
4 Step-Staircase / Grid Width 12
Suspected formula: T = 10n + 130
Prediction / Test: 10 x 15 + 130 = 280
15 + 16 + 17 + 18 + 27 + 28 + 29 + 39 + 40 + 51 = 280
Algebraic Proof:
n + (n+1) + (n+2) + (n+3) + (n+12) + (n+13) + (n+14) + (n+24) + (n+25) + (n+36) = 10n + 130
5 Step-Staircase
5 Step-Staircase / Grid Width 8
Suspected formula: T = 15n + 180
Prediction / Test: 15 x 34 + 180 = 690
34 + 35 + 36 + 37 + 38 + 42 + 43 + 44 + 45 + 50 + 51 + 52 + 58 + 59 + 66 = 690
Algebraic Proof:
n + (n+1) + (n+2) + (n+3) + (n+4) + (n+8) + (n+9) + (n+10) + (n+11) + (n+16) + (n+17) + (n+18) + (n+24) + (n+25) + (n+32) = 15n + 180
5 Step-Staircase / Grid Width 9
Suspected formula: T = 15n + 200
Prediction / Test: 15 x 39 + 200 = 785
39 + 40 + 41 + 42 + 43 + 48 + 49 + 50 + 51 + 57 + 58 + 59 + 66 + 67 + 75 = 785
Algebraic Proof:
n + (n+1) + (n+2) + (n+3) + (n+4) + (n+9) + (n+10) + (n+11) + (n+12) + (n+18) + (n+19) + (n+20) + (n+27) + (n+28) + (n+36) = 15n + 200
5 Step-Staircase / Grid Width 10
Suspected formula: T = 15n + 220
Prediction / Test: 15 x 23 + 220 = 565
23 + 24 + 25 + 26 + 27 + 33 + 34 + 35 + 36 + 43 + 44 + 45 + 53 + 54 + 63 = 565
Algebraic Proof:
n + (n+1) + (n+2) + (n+3) + (n+4) + (n+10) + (n+11) + (n+12) + (n+13) + (n+20) + (n+21) + (n+22) + (n+30) + (n+31) + (n+40) = 15n + 220
Here, I can see a pattern of a jump size of 20 so I suspect the next two formulas will be T = 15n + 240 and T = 15n + 260
5 Step-Staircase / Grid Width 11
Suspected formula: T = 15n + 240
Prediction / Test: 15 x 36 + 240 = 780
36 + 37 + 38 + 39 + 40 + 47 + 48 + 49 + 50 + 59 + 59 + 60 + 69 + 70 + 80 = 780
Algebraic Proof:
n + (n+1) + (n+2) + (n+3) + (n+4) + (n+11) + (n+12) + (n+13) + (n+14) + (n+22) + (n+23) + (n+24) + (n+33) + (n+34) + (n+44) = 15n + 240
5 Step-Staircase / Grid Width 12
Suspected formula: T = 15n + 260
Prediction / Test: 15 x 27 + 260 = 665
27 + 28 + 29 + 30 + 31 + 39 + 40 + 41 + 42 + 51 + 52 + 53 + 63 + 64 + 75 = 665
Algebraic Proof:
n + (n+1) + (n+2) + (n+3) + (n+4) + (n+12) + (n+13) + (n+14) + (n+15) + (n+24) + (n+25) + (n+26) + (n+36) + (n+37) + (n+48) = 15n + 260
The Final Formula
Now that I have found out all of the formulas of the different combinations of stair sizes and grid widths, I must now found the ultimate formula which connects these two variables together to add up to the total of the numbers of any staircase on any grid. So therefore if I enter 3 into the formula for the staircase size, 10 for the grid width size and 1 for the bottom left step number, the formula should accumulate the terms to make the total of 50.
Firstly,
To make this process simpler, I have broken this into 3 sections:
Finding the “n” term of the formula – n being the bottom left number of the staircase.
Finding the “g” term of the formula – g being the grid width.
Finding the number term of the formula – the number term appearing after the formulas found from finding out the grid part of the final formula.
Finding the “n” term of the formula
To find this part of the formula, I have taken the formulas of all of the staircase sizes of a grid width of 10 from my table of formulas:
1 ÷ 2 = 0.5 therefore = 0.5s² (s being the staircase size)
Here I have a linear sequence of 0.5 so therefore the next part of the formula is 0.5s.
So far I have the formula of:
TS = n(0.5s² + 0.5s)
Finding the “g” term of the formula
To find this part of the formula, I came up with a theory containing the relationship between the numbers in the staircase algebraically as follows:
I created this because the relationship between the “g” term of the algebra in each of the boxes and the grid width size is exactly the same as follows:
Between these two grids of numbers, only looking at 1 and 11 – n and n+g, they are exactly the same, with the grid width of 10, 1 + 10 is 11. This rule can be applied to any of the cells in the grid.
Now I am going to use the method of differences again to find this part of the final formula by only looking at the “g” part of each staircase size:
Here I have a constant difference of 1 so therefore it is cubic:
1 ÷ 6 = 1/6s³
I expect the next part to be a quadratic thus I need to use the method of differences again:
Here, instead of a quadratic sequence, I have a linear sequence of -1/6 for the next part of the formula.
So far, I have a formula of:
TS = n(0.5s² + 0.5s) + g(1/6s³-1/6s)
However the formula is not finished yet, I need to add the formula part from the number terms from the “g” formulas above in the previous section.
Finding the number term of the formula
To find this term, I will refer back to the grid of algebra defined cells (n, n+1, n+g, etc.) and extract the number term out to find the method of differences (again!)
After finding out the method of differences, it turns out to be exactly the same sequence for finding out the “g” term considering the differences between the staircase sizes of the number terms follow the same sequence as the “g” term!
So now, there is a constant difference of 1, therefore, 1 ÷ 6 = 1/6s³
Now, I would expect a quadratic but since the sequence is exactly the same as the “g” term formula, I am predicting an automatic linear sequence:
As predicted, the second part to the formula is -1/6.
The overall product is now: TS = n(0.5s² + 0.5s) + g(1/6s³ - 1/6s) + (1/6s³ - 1/6s)
To neaten the formula up a bit:
This seems ludicrous at first so I will now substitute the letters with numbers to prove this formula works.
With “n” being 1, “s” being 3 and “g” being 10, I should come up with a total step number of 50.
Staircase Size = 3, Grid width = 10, bottom left number = 1
1 + 2 + 3 + 11 + 12 + 21 = 50
So I have seen that this works fully and now that I have the overall formula connecting the staircase sizes, the grid widths and the bottom left number of the staircases together; the total of the numbers in any staircase of any size of any grid width can be worked out by using this formula:
