To Work Out General Formula for a 10x10 Grid
I then had to find out a formula for working out any step anywhere on a 10x10 grid.
y = 2 >> 3x + 11
3x + 33
y = 3 >> 6x + 44 x + 33
4x + 66 Diff. 11
y = 4 >> 16 x + 110 x + 44
5x + 110 Diff. 11
y = 5 >> 15x + 220 x + 55
6x + 165
y = 6 >> 21 x +385
Because there are 3 steps before I could find a similarity I had to use this expression to work out the ‘c’ in y=mx+c
ay³ + by² + cy + d
I then gathered up 4 equations to use this formula correctly
y = 2 >> 8a + 4b + 2c + d = 11 || equation 1
y = 3 >> 27a + 9b + 3c +d = 44 || equation 2
y = 4 >> 64a + 16b + 4c + d = 110 || equation 3
y = 5 >> 125 a + 25b + 5c + d = 220 || equation 4
I then did equation 3 – equation 2 , equation 4 – equation 2, and equation 5 – equation 2. I also noticed that working from d backwards u multiply by y to get the coefficients of the letters.
19a + 5b + c = 33 || equation 4 (multiplied by 6)
56a + 12b + 2c = 99 || equation 5 (multiplied by 3)
117a + 21b + 3c = 209 || equation 6 (multiplied by 2)
114a + 30b + 6c = 198 || equation 4
168a + 36b + 6c = 297 || equation 5
234a + 42b + 6c = 418 || equation 6
I then had to subtract equation 6 from equation 4 and equation 5 from equation 4.
120a + 12b = 220 || equation 7
54a + 6b = 99 || equation 8 (multiplied by 2 to make 6b = 12b)
108a + 12b = 198 || equation 9
Then subtract equation 9 from equation 7.
12a = 22 || divide each side by 12 to separate a
a = 1.83 or 1 5/6
I then substituted a into equation 7
120( 1 5/6) + 12b = 220
220 + 12b = 220 || both sides containing 220 cancel out
Therefore b = 0
I then substituted a and b into equation 4
114(1 5/6) + 30(0) + 6c = 198
209 + 6c = 198 (move 209 onto other side to place 6c on its own)
6c = 198 – 209
6c = -11
c = -1 5/6
I then substituted a b and c into equation 1
8(1 5/6) + 4(0) + 2(-1 5/6) + d = 11
14 2/3 – 3 2/3 + d = 11
11+ d = 11 (11 cancels out on both sides)
d = 0
So : a = 1 5/6 : b = 0 : c = -1 5/6 : d = 0
I can say that c = -a
Having found this I then took the coefficients of the formulas for the steps and placed them in a table. I noticed they were triangle numbers from Pascal’s triangle.
M >> 3
3
M >> 6 1
4
M >> 10 1
5
M >> 15 1
6
M >> 21
Because there were only 2 steps to find a similarity I knew I needed this equation
ax² + bx +c
I then began to substitute.
x = 2 >> a2² + b2 + c = 3
x = 3 >> a3² + b3 + c = 6
x = 4 >> a4² + b4 + c = 10
Rearranging them and simplifying them I found these equations.
4a + 2b + c = 3 || equation 1
9a + 3b + c = 6 || equation 2
16a + 4b + c = 10 || equation 3
I had to then subtract equation 1 from 3 and equation 2 from 1
12a + 2b = 7 || Equation 4
5a + b = 3 || Equation 5
I then had to multiply equation 5 by 2 to make b = 2b
12a + 2b = 6 || Equation 6
Then subtract
2a = 1
a = ½
I then had to substitute a into equation 5
5(½) +b = 3
2.5 + b = 3
b = ½
2 + 1 + c = 3
c = 0
I gathered the terms and found y = (1/2x² + 1/2x) x Reference Number + c
Gathering c I formed this equation.
y = (1/2x² + 1/2x) x Reference Number + (1 5/6x³ - 1 5/6 x)
This will work on anything on 10x10 grid
I then went on to extend my investigation further by finding relationships between the step size and position of the stair on an 8x8 Grid.
8X8 Grid
Now I had experience in working this out I was able to quickly find the formulas for the steps.
2 Step:
9
1 2
x + (x + 1) + (x + 8)
Therefore: 3x + 9
3 Step:
17
9 10
1 2 3
x + (x +1) + (x + 2) + (x + 8) + (x + 16)
Therefore: 6x + 36
4 step
25
17 18
9 10 11
1 2 3 4
x + (x + 1) + (x + 2) + (x + 3) + (x + 8) + (x + 9) + (x + 10) + (x + 16) + (x + 17) +
(x + 24)
Therefore: 10n + 90
5 Step:
33
25 26
17 18 19
9 10 11 12
1 2 3 4 5
x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 8) + (x + 9) + (x + 10) + (x + 11) +
(x + 16) + (x + 17) + (x + 18) + (x +24) + (x + 25) + (x + 32)
Therefore: 15n + 80
6 Step:
41
33 34
25 26 27
17 18 19 20
9 10 11 12 13
1 2 3 4 5 6
x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) + (x + 8) + (x + 9) + (x + 10) + (x + 11) + (x + 12) + (x + 16) + (x + 17) + (x + 18) + (x + 19) + (x + 24) + (x + 25) + (x + 26) +
(x + 32) + (x + 33) + (x + 40)
Therefore: 21x + 315
As one can see the x is the same for the 10 x 10 grid and the 8 x 8 grid.
As before I now need to work out a general formula for working out any stair placed anywhere on an 8 x 8 grid. I used the same way I used to work out an 10 x 10 grid general formula.
y = 2 >> 3x + 9
3x + 27
y = 3 >> 6x + 36 x + 27
4x + 54 9
y = 4 >> 10x + 90 x + 36
5n + 90 9
y = 5 >> 15x + 180 x + 45
6n + 135
y = 6 >> 21x + 315
As before in the 10 x 10 grid it took 3 steps to find a similarity which means I need a cubic expression of ay³ + by² + cy + d
y = 2 >> 8a + 4b + 2c + d = 9 || Equation 1
y = 3 >> 27a + 9b + 3c + d = 36 || Equation 2
y = 4 >> 64 a + 16b + 4c + d = 90 || Equation 3
y = 5 >> 125a + 25b + 5c + d = 180 || Equation 4
I then had to do equation 2 – equation 1, equation 3 – equation 1, and equation 4 – equation 1
19a + 5b + c = 27 || Equation 5 (multiplied by 6)
56a + 12b + 2c = 81 || Equation 6 (multiplied by 3)
117a + 21b + 3c = 171 || Equation 7 (multiplied by 2)
114a + 30b + 6c = 162 || Equation 8
168a + 36b + 6c = 243 || Equation 9
234a + 42b + 6c = 342 || Equation 10
Equation 10 – Equation 8 and Equation 9 – Equation 8
120a + 12b = 180 || Equation 11
54a + 6b = 81 || Equation 12 (multiplied by 2 to get 6b to = 12b)
108a + 12b = 162 || Equation 13
Equation 11 – Equation 13
12a = 18 || Divide each side by 12 to get a separated
a = 1.5
Substitute a into equation 11
120(1.5) + 12b = 180
180 + 12b = 180 || 180 on both sides cancel out
12b = 0
b = 0
Sub. a and b into equation 8
114 (1.5) + 30(0) + 6c = 162
171 + 6c = 162 || move 171 onto other side
6c = -9 || divide each side by 6
c = -1.5
Substitute into equation 1
8a + 4b + 2c +d = 9
8(1.5) + 4(0) + 2(-1.5) + d = 9
12 – 3 + d = 9
d = 0
Therefore a = 1.5 : b = 0 : c = 1.5 : and d = 0
As before b and d both equal 0 and c = -a
M >> 3
3
M >> 6 1
4
M >> 10 1
5
M >> 15 1
6
M >> 21
This is the same as the 10 x 10 grid therefore
y = (1/2x² + 1/2x) x Reference number + c
I then compiled both equation into a equation y= mx+c
Therefore
y = (1/2x² + 1/2x) x Reference Number + (1.5x³ - 1.5x) ||for an 8 x 8 Grid
To extend the investigation further I have decided to work out a formula for working out any step stair placed anywhere on any sized grid.
y = (1/2x² + 1/2x) x Reference Number + (1 5/6x³ - 1 5/6 x) || 10 x 10 Grid
y = (1/2x² + 1/2x) x Reference Number + (1.5x³ - 1.5x) || 8 x 8 Grid
I noticed that (1/2x² + 1/2x) is always around
Because of this I have also noticed the general formula which is:
(1/2x² + 1/2x) x Reference Number + (G+1 x³ – G+1 n)
6 6
Where x is step size and G is grid size
Having found this I then decided to test it on a 9 x 9 grid with 2 step stair.
Using my formula I calculated :
(1/2(2²) + ½(2) 1 + (9+1 (2³) – 9 +1 (2)
6 6
(2 + 1) 1 + ( (10 x 8) – 10 x 2) )
6 6
(3) 1 + (13 1/3 – 3 1/3)
3 + (10) = 13
To check that this is correct
10
1 2
10 + 1 + 2 = 13
This proves that all my formulas are correct.
Conclusion
In conclusion I have answered all the questions I needed to find out in my investigation and I was able to successfully extend the problem further and work out a general formula for any stair size placed anywhere on any sized grid.
I was also able to use my algebraic knowledge to carry out the investigation as well as substitution.
I could also extend the investigation further by working out a formula for stair sizes on 3 dimensional grids.
My results were justified by solving other problems using them and checking them by working it out the longer way.
Contents
Page 1: Introduction
Page 2: 3 step stair on a 10 x 10 grid and other step stairs on a 10 x 10 grid
Page 3: Other step stairs on a 10 x 10 Grid
Page 4: To Work Out General Formula for a 10x10 Grid
Page 5: To Work Out General Formula for a 10x10 Grid
Page 6: To Work Out General Formula for a 10x10 Grid
Page 7: To Work Out General Formula for a 10x10 Grid
Page 8: 8 x 8 Grid
Page 9: 8 x 8 Grid
Page 10: General Formula for an 8 x 8 Grid
Page 11: General Formula for an 8 x 8 Grid and Ultimate Formula
Page 12: Ultimate Formula
Page 13: Conclusion
