I have realised that the square above the stair-number is (n + 5) because of the grid size, which is 5. This explains why the square above the stair number in the 3-stair shape on the 10 x 10 grid is (n + 10) and the square above that is [n + (10 x 2)] or (n + 20).
Therefore I can redraw the 2-stair shape on a 5 x 5 grid in terms of ‘n’ and ‘g’.
The stair total for this 3-stair shape will now be
n + (n + 1) + (n + g)
= n + n + 1 + n + g
= 3n + g + 1.
Just as these individual values in each square would always be the same no matter where this 2-stair shape is translated around any size grid, so the formula Un = 3n + g + 1 for working out the stair total for any given stair number – where ‘n’ is the stair number, ‘g’ is the grid size, and ‘Un’ is the term which is the stair total – should also always be the same no matter where this stair shape is translated around any size grid.
To check if this theory is correct, I will first substitute ‘n’ for 7 on the same 5 x 5 grid, and then change the grid size to see if the formula still works.
Stair-number (n) = 7
The grid is still 5 x 5, so ‘g’ = 5.
Stair-total (Un) = 3n + g + 1
(found using = 3(7) + (5) + 1
the formula) = 21 + 5 + 1
= 27
Stair-total (found by adding) = 7 + 8 + 12 = 27 ✓
This shows that my formula must work for all 2-stair numbers on the 5 x 5 grid.
However I was yet to find out if this formula would also work on other grid sizes. To investigate this, I am going to see if the formula still works for 2-stair shapes on an 8 x 8 grid.
Grid: 8 x 8
‘n’ = 46, ‘g’ = 8:
Stair-total (Un) = 3n + g + 1
(found using = 3(46) + (8) + 1
the formula) = 138 + 9
= 147
Stair-total = 46 + 47 + 54
(found by = 147 ✓
adding)
This must mean that my formula Un = 3n + g + 1 – where ‘n’ is the stair number, ‘g’ is the grid size, and ‘Un’ is the term which is the stair total – works for all 2-stair shapes on any size grid.
Next I will find a formula for calculating the stair-total of a 3-stair shape for any given stair-number. So far I only know the formula for working out the stair-total of 3-stair shape for a 10 x 10 grid.
Stairs: 3
To find a formula for calculating the stair total for any given stair-number of a 3-stair shape on a 5 x 5 grid, I will draw out the 3-stair shape on a 5 x 5 grid in terms of ‘n’ and ‘g’.
Grid: 5 x 5
The stair total for this 3-stair shape is
n + (n + 1) + (n + 2) + (n + g) + (n + g + 1) + (n + 2g)
= n + n + 1 + n + 2 + n + g + n + g + 1 + n + 2g
= 6n + 4g + 4.
As said before, just as these individual values in each square would always be the same no matter where this 3-stair shape is translated around any size grid, the formula Un = 6n + 4g + 4 for working out the stair total for any given stair number – where ‘n’ is the stair number, ‘g’ is the grid size, and ‘Un’ is the term which is the stair total – would also always be the same no matter where this stair shape is translated around any size grid.
As a check to support this theory, I will substitute ‘n’ for 11 on the same 5 x 5 grid.
Stair-number (n) = 11
The grid is still 5 x 5, so ‘g’ = 5.
Stair-total (Un) = 6n + 4g + 4
(found using = 6(11) + 4(5) + 4
the formula) = 66 + 20 + 4
= 90
Stair-total (found by adding) = 11 + 12 + 13 + 16 + 17 + 21 = 90 ✓
This shows that my formula must work for all 3-stair shapes on the 5 x 5 grid.
To investigate if this formula also works for 3-stair shapes on other grid sizes, I am going to see if the formula still works for 3-stair shapes on a 10 x 10 grid.
Grid: 10 x 10
‘n’ = 52, ‘g’ = 10:
Stair-total (Un) = 6n + 4g + 4
(found using = 6(52) + 4(10) + 4
the formula) = 312 + 40 + 4
= 356
Stair-total = 52 + 53 + 54 + 62 + (found by 63 + 72
adding)
= 356 ✓
This must mean that my formula Un = 6n + 4g + 4 – where ‘n’ is the stair number, ‘g’ is the grid size, and ‘Un’ is the term which is the stair total – works for all 3-stair numbers on any size grid.
Something I have noticed is that because this formula Un = 6n + 4g + 4 works for 3-stair shapes on a 10 x 10 grid, it also confirms the formula Un = 6n + 44 for 3-stair shapes on 10 x 10 grids, which I had previously found.
g = 10 (because the grid is 10 x 10)
Un = 6n + 4g + 4
= 6n + 4(10) + 4
= 6n + 40 + 4
= 6n + 44
Next I will find a formula for calculating the stair-total of a 4-stair shape for any given stair-number.
Stairs: 4
To find a formula for calculating the stair total for any given stair-number of a 4-stair shape on a 5 x 5 grid, I will draw out the 4-stair shape on a 5 x 5 grid in terms of ‘n’ and ‘g’.
Grid: 5 x 5
The stair total for this 3-stair shape is
n + (n + 1) + (n + 2) + (n + 3) + (n + g) + (n + g + 1) + (n + g + 2) + (n + 2g) + (n + 2g + 1) + (n + 3g)
= n + n + 1 + n + 2 + n + 3 + n + g + n + g + 1 + n + g + 2 + n + 2g + n + 2g + 1 + n + 3g
= 10n + 10g + 10.
I think that the formula Un = 10n + 10g + 10 for working out the stair total for any given stair number – where ‘n’ is the stair number, ‘g’ is the grid size, and ‘Un’ is the term which is the stair total – would also always be the same no matter where this stair shape is translated around any size grid.
As a check to support this theory, I will substitute ‘n’ for 6 on the same 5 x 5 grid.
Stair-number (n) = 6
The grid is still 5 x 5, so ‘g’ = 5.
Stair-total (Un) = 10n + 10g + 10
(found using = 10(6) + 10(5) + 10
the formula) = 60 + 50 + 10
= 120
Stair-total (found by adding) = 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 16 + 17 + 21 = 120 ✓
This shows that my formula must work for all 4-stair shapes on the 5 x 5 grid.
To investigate if this formula also works for 4-stair shapes on other grid sizes, I am going to see if the formula still works for 4-stair shapes on a 7x 7 grid.
Grid: 7
‘n’ = 22, ‘g’ = 7:
Stair-total (Un) = 10n + 10g + 10
(found using = 10(22) + 10(7) + 10
the formula) = 220 + 70 + 10
= 300
Stair-total = 22 + 23 + 24 + 25 + 29 + 30 + 31 + 36 (found by 37 + 43
adding)
= 300 ✓
This must mean that my formula Un = 10n + 10g + 10 – where ‘n’ is the stair number, ‘g’ is the grid size, and ‘Un’ is the term which is the stair total – works for all 4-stair numbers on any size grid.
Next I will find a formula for calculating the stair-total of a 5-stair shape for any given stair-number.
Stairs: 5
To find a formula for calculating the stair total
for any given stair-number of a 5-stair shape I will use an 8 x 8 grid because there might not be enough room to check a formula by translations later on, on a 5 x 5 grid. I will draw out the 5-stair shape on an 8 x 8 grid in terms of ‘n’ and ‘g’.
Grid: 8 x 8
The stair total for this 5-stair shape is
n + (n + 1) + (n + 2) + (n + 3) + (n + 4) + (n + g) + (n + g + 1) + (n + g + 2) + (n + g + 3) + (n + 2g) + (n + 2g + 1) + (n + 2g + 2) + (n + 3g) + (n + 3g + 1) + (n + 4g)
= n + n + 1 + n + 2 + n + 3 + n + 4 + n + g + n + g + 1 + n + g + 2 + n + g + 3 + n + 2g + n + 2g + 1 + n + 2g + 2 + n + 3g + n + 3g + 1 + n + 4g
= 15n + 20g + 20.
I think that the formula Un = 15n + 20g + 20 for working out the stair total for any given stair number – where ‘n’ is the stair number, ‘g’ is the grid size, and ‘Un’ is the term which is the stair total – would also always be the same no matter where this stair shape is translated around any size grid.
As a check to support this theory, I will substitute ‘n’ for 19 on the same 8 x 8 grid.
Stair-number (n) = 19
The grid is still 8 x 8, so ‘g’ = 8.
Stair-total (Un) = 15n + 20g + 20
(found using = 15(19) + 20(8) + 20
the formula) = 285 + 160 + 20
= 465
Stair-total (found by adding) = 19 + 20 + 21 + 22 + 23 + 27 + 28 + 29 + 30 + 35 + 36 + 37 + 43 + 44 + 51 = 465 ✓
This shows that my formula must work for all 5-stair shapes on the 8 x 8 grid.
To investigate if this formula also works for 5-stair shapes on other grid sizes, I am going to see if the formula still works for 5-stair shapes on a 9 x 9 grid.
Grid: 9 x 9
‘n’ = 23, ‘g’ = 9:
Stair-total (Un) = 15n + 20g + 20
(found using = 15(23) + 20(9) + 20
the formula) = 345 + 180 + 20
= 545
Stair-total = 23 + 24 + 25 + 26 + (found by 27 + 32 + 33 + 34 + adding) 35 + 41 + 42 + 43 +
50 + 51 + 59
= 545 ✓
This must mean that my formula Un = 15n + 20g + 20 – where ‘n’ is the stair number, ‘g’ is the grid size, and ‘Un’ is the term which is the stair total – works for all 4-stair numbers on any size grid.
Finally, I will find a formula for calculating the stair-total of a 6-stair shape for any given stair-number.
Stairs: 6
To find a formula for calculating the stair total for any given stair-number of a 6-stair shape I will use an 8 x 8 grid. I will draw out the 6-stair shape on an 8 x 8 grid in terms of ‘n’ and ‘g’.
Grid: 8 x 8
The stair total for this 6-stair shape is
n + (n + 1) + (n + 2) + (n + 3) + (n + 4) + (n + 5) + (n + g) + (n + g + 1) + (n + g + 2) + (n + g + 3) + (n + g + 4) + (n + 2g) + (n + 2g + 1) + (n + 2g + 2) + (n + 2g + 3) + (n + 3g) + (n + 3g + 1) + (n + 3g + 2) + (n + 4g) + (n + 4g + 1) + (n + 5g)
= n + n + 1 + n + 2 + n + 3 + n + 4 + n + 5 + n + g + n + g + 1 + n + g + 2 + n + g + 3 + n + g + 4 + n + 2g + n + 2g + 1 + n + 2g + 2 + n + 2g + 3 + n + 3g + n + 3g + 1 + n + 3g + 2 + n + 4g + n + 4g + 1 + n + 5g
= 21n + 35g + 35.
I think that the formula Un = 21n + 35g + 35 for working out the stair total for any given stair number – where ‘n’ is the stair number, ‘g’ is the grid size, and ‘Un’ is the term which is the stair total – would also always be the same no matter where this stair shape is translated around any size grid.
As a check to support this theory, I will substitute ‘n’ for 11 on the same 8 x 8 grid.
Stair-number (n) = 11
The grid is still 8 x 8, so ‘g’ = 8.
Stair-total (Un) = 21n + 35g + 35
(found using = 21(11) + 35(8) + 35
the formula) = 231 + 280 + 35
= 546
Stair-total (found by adding) = 11 + 12 + 13 + 14 + 15 + 16 + 19 + 20 + 21 + 22 + 23 + 27 + 28 + 29 + 30 + 35 + 36 + 37 + 43 + 44 + 51 = 546 ✓
This shows that my formula must work for all 6-stair shapes on the 8 x 8 grid.
To investigate if this formula also works for 6-stair shapes on other grid sizes, I am going to see if the formula still works for 6-stair shapes on a 9 x 9 grid.
Grid: 9 x 9
‘n’ = 23, ‘g’ = 9:
Stair-total (Un) = 21n + 35g + 35
(found using = 21(22) + 35(9) + 35
the formula) = 462 + 315 + 35
= 812
Stair-total = 22+ 23 + 24 + 25 + 26 +
(found by 27 + 31 + 32 + 33 + 34 + adding) 35 + 40 + 41 + 42 + 43 +
49 + 50 + 51 + 58 + 59 + 67
= 812 ✓
This must mean that my formula Un = 21n + 35g + 35 – where ‘n’ is the stair number, ‘g’ is the grid size, and ‘Un’ is the term which is the stair total – works for all 6-stair numbers on any size grid.
Next I will try and find a general rule for finding the stair total for an n-stair shape on an n x n grid.
As a recap of all these results I have obtained so far, I will draw up a table:
I have added the formula for working out the stair total for a one-stair shape. This stair total is just the stair number as there is just one stair, which is ‘n’. Although this is very obvious, it is important that I complete the table as it will facilitate the process of finding a formula later on.
I have noticed that all the formulae are made up of three terms: an ‘n’ term, a ‘g’ term and a ‘number’ term (the first formula could be written as 1n + 0g + 0). These terms are always added together to form the formula. The coefficient of ‘g’ is always the same as the number in the ‘number’ term.
Therefore I can tackle this problem of finding a rule connecting the number of stairs and its formula for working out the stair total in two parts: finding the rule connecting the number of stairs and the coefficient of the ‘n’ term, and then finding a rule connecting the number of stairs and the coefficient of the second (and in effect the third term as well).
Hence, firstly, I will write out the sequence where I am going to put the number of stairs as the position and the ‘n’ term as the term for the sequence:
I have realised that this sequence is made up of the triangle numbers. From previous work I know that the formula for term ‘t’ in the sequence is ½ t (t + 1).
Therefore the 7th term will be ½ (7) [(7) + 1] = ½ 7 (8) = ½ x 56 = 28
To check this I will add up the numbers that form a triangle with 7 rows:
1 + 2 + 3 + 4 + 5 + 6 + 7 = 28.
Hence, the formula for working out the stair total of a 7-stair shape will start like this: 28n + …
As there is a 3rd difference in this sequence, this means that it is a cubic equation.
Therefore, if I use the cubic equation ax3 + bx2 + cx + d, I will be able to find the 4 unknowns.
Term 1: a(1) 3 + b(1) 2+ c(1) + d = 0
a + b + c + d = 0 -------- ①
Term 2: a(2)3 + b(2)2 + c(2) + d = 1
8a + 4b + 2c + d = 1 -------- ②
Term 3: a(3)3 + b(3)2 + c(3) + d = 4
27a + 9b + 3c + d = 4 -------- ③
Term 4: a(4)3 + b(4)2 + c(4) + d = 10
64a + 16b + 4c + d = 10 -------- ④
Term 4-term 3: 37a + 7b + c = 6 … this will become number 5
Term 3- term 2: 19a + 5b +c = 3… this will become number 6
Term 2- term1: 7a + 3b + c =1… This will become number 7
Term 5 – term 6: 18a + 2b = 3 … this will become number 8
Term6 – term 7: 12a + 2b = 2 …This will become number 9
Term 8- term 9: 6a = 1
A = 1/6
Therefore term 8 becomes 3 + 2b = 3
Which means that b = 0
In term 7: 7/6 + c = 1
Which means that C = -1/6
In term 1 A and C cancel each other out, and with B equalling zero it must mean that D also equals zero.
Therefore the expression is 1/6x3 – 1/6x
e.g. x = 3
27/6 – ½ = 4
This fits in with the above table, to make sure another number will be substituted
X = 5
125/6 – 5/6 = 120/6 = 20
Again this fits into the table meaning that my formula is correct.
