Stair 1:
9 + 10 + 11 + 17 + 18 + 25 = 90
F+F+1+F+2+F+8+F+8+1+F+16=
6F+36=90
6F=90-36=54
6F=54÷9=6
F=9
6(9) + 36
Stair 2:
22 + 23 + 24 + 30 + 31 + 38 = 168
G+G+1+G+2+G+8+G+8+1+G+16=
6G+36=168
6G=168-36=132
6G=132÷22=6
G=22
6(22) + 36
Stair 3:
46 + 47 + 48 + 54 + 55 + 62=312
H+H+1+H+2+H+8+H+8+1+H+16=
6H+36=312
6H=312-36=276
6H=276÷46=6
H=46
6(46) + 36
Stair 4:
41 + 42 + 43 + 49 + 50 + 57 = 282
I+I+1+I+2+I+8+I+8+1+I+16=
6I+36=282
6I=282-36=246
6I=246÷41=6
I=41
6(41) + 36
7X7 Grid
Stair 1:
19 + 20 + 21 + 26 + 27 + 33 = 146
J+J+1+J+2+J+7+J+7+1+J+14=
6J+32=146
6J=146-32=114
6J=114÷19=6
J=19
6(19) + 32
Stair 2:
30 + 31 + 32 + 37 + 38 + 44 = 212
K+K+1+K+2+K+7+K+7+1+K+14=
6K+32=212
6K=212-32=180
6K=180÷30=6
K=30
6(30) + 32
Stair 3:
15 + 16 + 17 + 22 + 23 + 29 = 122
L+L+1+L+2+L+7+L+7+1+L+14=
6L+32=122
6L=122-32=90
6L=90÷15=6
L=15
6(15) + 32
Stair 4:
4 + 5 + 6 + 11 + 12 + 18 = 56
#
M+M+1+M+2+M+7+M+7+1+M+14=
6M+32=56
6M=56-32=24
6M=24÷4=6
M=4
6(4) + 32
6X6 Grid
Stair 1:
1 + 2 + 3 + 7 + 8 + 13 = 34
N+N+1+N+2+N+6+N+6+1+N+12=
6N+28=34
6N=34-28=6
6N=6÷1=6
N=6
6(1) + 28
Stair 2:
4 + 5 + 6 + 10 + 11 + 16 = 52
O+O+1+O+2+O+7+O+7+1+O+12=
6O+28=52
6O=52-28=24
6O=24÷4=6
6O=4
6(4) + 28
Stair 3:
19 + 20 + 21 + 25 + 26 + 31 = 142
P+P+1+P+2+P+7+P+7+1+P+12=
6P+28=142
6P=142-28=114
6P=114÷19=6
P=19
6(19) + 28
Stair 4:
22 + 23 + 24 + 28 + 29 + 34 = 160
S+S+1+S+2+S+7+S+7+1+S+12
6S+28=160
6S=160-28=132
6S=132÷22=6
S=22
6(22) + 28
What Next?
Now that I have produced 3 step stairs on 10X10, 9X9, 8X8, 7X7 and on 6X6 grids sized, I can now put my results and transfer them onto a table. I have chosen to show my algebraic formulae on a table as it will be easier for me to read of my rules as to which I have worked out. On my table I will show the general formula and the algebra of a particular step stair of a grid for each of the different sized grids which have been displayed.
From my table of results on the previous page, I have noticed a general trend:
I have noticed that as that grid size depreciates e.g. by one size from a 10X10 to a 9X9, the algebra also decreases by 4. This is because the grid size is getting smaller and therefore the numbers inside the grid are being reduced from 100 to 81.
- I will now test my theory that the size of the grid will depend the outcome of the algebra and the algebra will be 4 less then its predecessor e.g. the algebra changing from 6n + 28 on a 6X6 grid to 6n + 20 on a 4X4 grid. I will prove this by drawing up a 4X4 grid. The ‘Letter’ value in the grid is ‘n’.
1 + 2 + 3 + 5 + 6 + 9 = 26
n+n+1+n+2+n+4+n+4+1+n+8=
6n+20=26
6n=26-20=6
6n=6÷1=6
n=1
6(1) + 20
My test worked. It shows that the size of the grid always shows the outcome of the algebra and most importantly I have noticed the trend which has reduced to 20 as I predicted.
Extending my Original line of Enquiry:
Now that I have finished my 3 step stairs on different grid sizes, I will now extend my line of enquiry by investigating the relationship between the stair total of 4 step stairs, 5 step stairs and 2 step stairs all of which will be produced on 10X10 grids to see how the stair totals differ from one and other and how their basic algebra differentiates.
4 Step Stair on a 10X10 Grid
Stair 1:
61 + 62 + 63 + 64 + 71 + 72 + 73 + 81 + 82 + 91 = 721
N+N+1+N+2+N+3+N+10+N+10+1+N+10+2+N+20+N+21+N+30 =
10N+110=721
10N=721-110=611
10N=611÷61=10
N=61
61(10) + 110
Stair 2:
67 + 68 + 69 + 70 + 77 + 78 + 79 + 87 + 88 + 97 = 780
X+X+1+X+2+X+3+X+10+X+10+1+X+10+2+X+20+21+X+30=
10X+110=780
10X=780-110=670
10X=670÷67=10
X=67
67(10) + 110
Stair 3:
1 + 2 + 3 + 4 + 11 + 12 + 13 + 21 + 22 + 31 = 120
12
Y+Y+1+Y+2+Y+3+Y+10+Y+10+1+Y+10+2+Y+20+Y+21+Y+30=
10Y+110=120
10Y=120-110=10
10Y=10÷1=10
Y=1
1(10) + 110
Stair 4:
6 + 7 + 8 + 9 + 16 + 17 + 18 + 26 + 27 + 36 = 170
Z+Z+1+Z+2+Z+3+Z+10+Z+10+1+Z+10+2+Z+20+Z+21+Z+30=
10Z+110=170
10Z=170-110=60
10Z=60÷6=10
Z=6
6(10) + 110
Now that I have completed the 4 step stair on a 10X10 sized grid, I will now test to see if the formula which I have found (10x+110) will work on a 4 step stair which is turned the opposite direction. I will test to see if the formula that I have found out will still work on a 4 step stair of an example from above. I have chosen to use stair 4.
6 + 7 + 8 + 9 + 16 + 17 + 18 + 26 + 27 + 36 = 170
Z+Z-1+Z-2+Z-3+Z+10+Z+10-1+Z+10-2+Z+20+Z+20-1+Z+30 =
10Z+110=170
10Z=170-110=60
10Z=60÷6=10
Z=6
6(10) + 110
I have found out through this worked example that by rotating the 4 step stair or turning it to the left, the same rule will be formed. The rule will be 6x10+110. This proves that my theory worked and that if you keep the same numbers in the number stair and rotate the direction of the stair to the opposite direction then the same stair total will be formed 170 and the same amount of numbers added up in the formula – 110. So as a result you will be still working with the same numbers and end up doing the same calculations. But if you were to change the numbers in the stair around by using different numbers, then you will be playing with a different stair total and therefore the rule will be different.
2 Step Stair on a 10X10 Grid
Stair 1:
74 + 75 + 84 = 233
A+A+1+A+10=
3A+11=233
3A=233-11=222
3A=222÷74
A=74
74(3) + 11
Stair 2:
51 + 52 + 61 = 164
B+B+1+B+10=
3B+11=164
3B=164-11=153
3B=153÷51
B=3
51(3) + 11
Stair 3:
78 + 79 + 88 = 245
C+C+1+C+10=
3C+11=245
3C=245-11=234
3C=234÷78=3
C=78
78(3) + 11
Stair 4:
39 + 40 + 49 = 128
D+D+1+D+10=
3D+11=128
3D=128-11=117
3D=117÷39=3
D=39
39(3) + 11
Stair 5:
44 + 45 + 54 = 143
E+E+1+E+10=
3E+11=143
3E=143-11=132
3E=132÷44=3
E=44
44(3) + 11
5 Step Stair on a 10X10 Grid
Stair 1:
1 + 2 + 3 + 4 + 5 + 11 + 12 + 13 + 14 + 21 + 22 + 23 + 31 + 32 + 41 = 235
A+A+1+A+2+A+3+A+4+A+10+A+10+1+A+10+2+A+10+3+A+20+A+21+A+22+A+30+A+31+A+40=
15A+220=235
15A=235-220=15
15A=15÷1=15
A=1
1(15) + 220
Stair 2:
6 + 7 + 8 + 9 + 10 + 16 + 17 + 18 + 19 + 26 + 27 + 28 + 36 + 37 + 46 = 310
B+B+1+B+2+B+3+B+4+B+10+B+10+1+B+10+2+B+10+3+B+20+B+21+B+22+B+30+B+31+B+40=
15B+220=310
15B=310-220=90
15B=90÷6=15
B=6
6(15) + 220
Stair 3:
51 + 52 + 53 + 54 + 55 + 61 + 62 + 63 + 64 + 71 + 72 + 73 + 81 + 82 + 91 = 985
PTO – Please Turn Over
C+C+1+C+2+C+3+C+4+C+10+C+10+1+C+10+2+C+10+3+C+20+C+21+C+22+C+30+C+31+C+40=
15C+220=985
15C=985-220=765
15C=765÷51=15
C=51
51(15) + 220
Stair 4:
56 + 57 + 58 + 59 + 60 + 66 + 67 + 68 + 69 + 76 + 77 + 78 + 86 + 87 + 96 = 1060
D+D+1+D+2+D+3+D+4+D+10+D+10+1+D+10+2+D+10+3+D+20+D+21+D+22+D+30+D+31+D+40=
15D+220=1060
15D=1060-220=840
15D=840÷56=15
D=56
56(15) + 220
After drawing up 4 5 step stairs on a 10X10 grid and finding a formula (15x+220) I will now test to see if the rule that I have found out works when the 5 step stair is turned around. I will be using all the same numbers from inside the step stair in order to run this test. I will use example 4 and predict that the same rule will work on the step stair.
56 + 57 + 58 + 59 + 60 + 66 + 67 + 68 + 69 + 76 + 77 + 78 + 86 + 87 + 96 = 1060
D+D+1+D+2+D+3+D+4+D+10+D+10+1+D+10+2+D+10+3+D+20+D+21+D+22+D+30+D+31+D+40=
15D+220=1060
15D=1060-220=840
15D=840÷56=15
D=15
56(15) + 220
By looking at the test above, I can clearly see that having turned around the 5 step stair has affected the original formula and also more importantly, it proves and shows that my theory worked that the formula will work on the rotated step stair. The same formula will work on the 5 step stair as the numbers inside the step stair have not changed in value have not been rearranged so as a result I am dealing with the same numbers but ones that are swapped or moved to the opposite direction.
6 Step Stair on a 10X10 Grid
Stair 1:
41 + 42 + 43 + 44 + 45 + 46 + 51 + 52 + 53 + 54 + 55 + 61 + 62 + 63 + 64 + 71 + 72 + 73 + 81 + 82 + 91= 1246
L+L+1+L+2+L+3+L+4+L+5+L+10+L+10+1+L+10+2+L+10+3+L+10+4+L+20+L+
21+L+22+L+23+L+30+L+31+L+32+L+40+L+41+L+50=
21L+385=1246
21L=1246-385=861
21L=861÷41=21
L=41
41 (21) + 385
Moving on, I have found a formula which works out how to calculate the stair total for a 6 step stair. The formula is 21X+385. I got this formula when I worked it out because if you add up all the numbers inside the step stair on the right hand side of the step stair, the total is 385. I got 21L because there are 21 L’s inside the step stair on the right hand side. Having now known this, I will now attempt to see if this formula will work when the 6 step stair is rotated in the opposite direction and will use the same numbers inside the step stair. I predict that when I do this the same formula will work and there will be no need to find a new one nor one will be formed as I will be still using the same numbers and the same step stair. I will use stair 1 to demonstrate this.
41 + 42 + 43 + 44 + 45 + 46 + 51 + 52 + 53 + 54 + 55 + 61 + 62 + 63 + 64 + 71 + 72 + 73 + 81 + 82 + 91 = 1246
L+L+1+L+2+L+3+L+4+L+5+L+10+L+10+1+L+10+2+L+10+3+L+10+4+L+20+L+
21+L+22+L+23+L+30+L+31+L+32+L+40+L+41+L+50=
21L+385=1246
21L=1246-385=861
21L=861÷41=21
L=41
41 (21) + 385
From looking at the evidence above, it shows testing and shows that what I predicted has come true. Even though I have turned the step stair around I am still using the same numbers so as a result the numbers in the ‘L’ boxes add up to 21L and if you add up all the numbers inside it, they add up to 385. So even though I have turned the step stair around, this has not affected the values inside the step stair. All it has done is display the values in a different view.
What Next?
Now that I have completed my 3 step stairs on a 10X10, 9X9, 8X8, 7X7 and on a 6X6 grid, completed a series of 2 Step stairs, 4 step stairs, 5 step stairs and 6 step stairs and found their formulas, I will now be able to gather the general formulas of each step stair on a 10X10 grid and show it on a table. I have chosen to show this data on a table as it will be easier to notice and read off my values.
Table of Formulas
From looking at the table above, it shows that the intervals between each set of formulas are increasing. The table also shows that each time the step stair increases by 1. To calculate the formula for the next step stair in the table for example, if you wanted to find the formula of a 3 step stair from a 2 step stair, you add 3 to the formula to make it 6x. To find the first number from a 4 step stair to a 5 step stair you add 5 to make it 15x.
The common difference for these formulas is +20. So, as a result the equation or rule is:
n = 15n +20g + 20
The difference between the ‘n ‘ numbers are :
3n 6n 10n 15n
1st Difference
2nd Difference
The 2nd difference is the same; therefore it will be solved using a quadratic equation simultaneously. A x² + bx + c. This is the formula for a quadratic equation.
To find the values of A, B and C we do this…….
A x² + bx + c
X = 1: A + B + C = 1
X = 2: 4a + 2B + C = 3 We Substitute our values into
the quadratic equation:
X = 3: 9a + 3B + C = 6
To get a:
4a + 2b + c = 3 9a + 3b + c = 6
a + b + c = 1 4a + 2b + c = 3
3a + b = 2
5a + b = 3
3a + B = 2
2a = 1
a = ½
To get b:
3a + b = 2
3 x ½ + b = 2
1 ½ + b = 2 2 – 1 ½ = ½ = b
b = ½
To get C:
We put ‘a’ and ‘b’ into the x=1 equation: a + b + c = 1
We now substitute our terms into the original formula:
A x² + Bx + C
a = ½ ½ X² + ½ X + 0
b = ½
c = 0 ½ X² + ½ X
This also equals the rule for triangle numbers, ½ n² + ½n
Triangle Numbers:
1 3 6 10 15 21 28 36 45 55…
Triangle numbers are numbers where the difference is an increasing set of differences. For example:
Having blobs in the shape of triangles where in every new triangle the blob has increased by 1. The number sequence is shown, with the differences increasing by 1 each time.
The formula for triangle numbers is:
nth term = ½ n (n+1), this is the same as my formula when ‘n’ and ‘x’ mean the same thing:
½ X (x+1)
Testing my formula:
½ x² + ½ x = ½ X (x+1)
when x = 4, ½ + 16 + ½ X 4 =10
8 + 2 = 10
Pyramid Numbers
The 3rd differences are the same, so as a result it is cubic. The cubic formula is:
a X³ + b X ² + c X + D
Substitute:
1x = 1: A + B + C + D = 0
2x = 2: 8A + 4B + 2D + D = 1
3x = 3: 27A + 9B + 3C + D =4
4x = 4: 64A + 16B + 4C + D = 10
8A + 4B + 2C + D = 1 64A + 16B + 4C + D = 10
A + B + C + D = 0 27A + 9B + 3C + D = 4
7A + 3B + c = 1 37A + 7B + C = 6
eliminates D: 27A + 9B + 3C + D = 4
8A + 4B + 2C + D = 1
19A + 5B + C = 3
19A + 5B + C = 3 37A + 7B + C = 6
7A + 3B + C = 1 eliminates C 19A + 5B + C = 3
12A + 2B = 2 18A + 2B = 3
to get B:
18A + 2B = 3 substitute: 12A + 2B = 2
12A + 2B = 2 a = ¹/6 12x ¹/6 + B = 2
6A = 1 12x ¹/6 + 0 = 2
A = ¹/6 B = 0
to get C: now we put our A, B + C values
into the cubic number formula:
Substitute: 7A + 3b + C = 1 a = ¹/6: a X ³ + b + x² + CX + d
A = ¹/6 7x ¹/6 + 3x D + C = 1 b = 0 : ¹/6 x³ + - ¹/6X + 0
B = ¹/6 1¹/6 + D + C = 1 c = - ¹/6: ¹/6 X³ + - ¹/6 X
d = 0
Now I am going to test my findings on my general rule, which is:
(½ ² + ½ ) n + (¹/6 X² = ¹/6x) g+ (¹/6 X² - ¹/6X) + (½ X 4² + ½ 4) + ¼ X 4 = 60
x = 4
n = 1
g = 4
My general rule for any grid size with any number of stairs has been proven to work.
Why you get triangle numbers and Why you get pyramid numbers:
1 Step Stair = 1n, 2 Step Stair = 3n + g + 1, 3 Step Stair = 6n + 6g + 4, 4 Step Stair = 10n +10g + 10
