For other 3-step stairs, investigate the relationship between the stair total and the position of the stair shape on the grid. To start the investigation a 10x10-numbered grid square is used as illustrated below in table 1:
Mill Hill County High School
Year 11
Mathematics GCSE Coursework
EDEXCEL 2003, SYLLABUS 1387/1388
F, I & H
Tejesh Patel
Class 11H
Assignment Part 1
Below is a 10x10 number grid:
The total on the numbers coloured in blue = 90 (i.e. 1+11+12+21+22+23)
Therefore the stair total in this 3-step stair = 90
Part 1 Objective:
For other 3-step stairs, investigate the relationship between the stair total and the position of the stair shape on the grid.
To start the investigation a 10x10-numbered grid square is used as illustrated below in table 1:
From the complete 10 x 10 numbered grid square, we use part of it to carry out our initial investigation, for example the grid box on the right shows a slice of the 10 x 10 gird square, i.e. 6 boxes representing the numbers 1,11,12,21,22,23 (The 3-step stair)
From this basic numbered square that looks like stairs or steps we can start to establish if there is a pattern. If a pattern is found then we can use an algebra equation to represent this pattern and use the equation for a 10 x 10 numbered Grid Square.
By using the 3-step stair example we know there are [6] squares and lets assume in a 3-step stair the bottom grid box is equal to [x], therefore in our 3-step stair x = 21
Using the values in algebra the formula(s) would look like this:
The 1st square = 1 then the formula is x - 20 = 1
The 2nd square = 11 then the formula is x - 10 = 11
The 3rd square = 12 then the formula is x - 9 = 12
The 4th square = 21 it is simply just x = 21
The 5th square = 22 then the formula is x + 1 = 22
The 6th square = 23 then the formula is x + 2 = 23
The above algebra equations are shown below in our 3-step stair:
x - 20
x - 10
x - 11
X
x + 1
x + 2
By adding the values from the equation (-20 - 10 - 9 + 1 + 2) = [-36]
Thus we can use the algebra equation 6x - 36 = [the total value of squares in a 10x10 square]
Using the above logic and method we can use it in other grids, such as an 11x11 and a 12x12 numbered grid square. Using the same 3-step stair approach we can use the theory for the 11x11 and 12x12-numbered square to find a pattern.
Below are the results from the such an exercise on a 10x10, 11x11 and 12x12 numbered grid box
Table 4: 10 x 10 Grid
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
00
Table 5: Using the algebra equation and adding the values we get -36
x - 20
x - 10
x - 11
X
x + 1
x + 2
Table 6: 11 x 11 Grid
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
00
01
02
03
04
05
06
07
08
09
10
Table 7:
x - 32
x - 11
x - 10
X
x + 1
X + 2
Using the algebra equation and adding the values we get -40
Table 8: 12 x 12 Grid
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
00
01
02
03
04
05
06
07
08
09
10
11
12
13
14
15
16
17
18
19
20
Table 9:
x - 24
x - 12
x - 11
x
x + 1
x + 2
Using the algebra equation and adding the values we get -44
If we closely examine the results from the 3-step stairs from the 3 numbered grid squares, i.e. 10x10, 11x11 and 12x12 we can see there is a constant number that is consistent, which is [4]
We can draw up a table by using this constant number as illustrated below in table 10:
Table 10:
From the table on the left and using our results from the investigation we can see a pattern emerging. Every time the size of the grid box increases by 1, the value in the algebra formula increases by 4. For example for a 10x10 numbered grid using a 3-step stair the formula is 6x-36 where x is the number in the bottom left hand square, then we increase the grid size by 1, 11x11 and using the same 3-stepped stair approach the formula is 6x-40, etc. We can clearly see the constant number [4] is consistent every time the grid size increases by [1].
For any 3-step numbered gird box as shown in table 2 or in the six examples below, and [x] as the value from the bottom left hand square and the algebra theory used to calculate the equation the total value of the squares can be found. This theory has been put to the test using a 10x10, 11x11 and 12x12 gird squares. The results are conclusive and consistent, proving the theory to be accurate and reliable.
Using any of the algebra equation from the above table, e.g. 6x-36 or 6x-44 we can prove the results for any 3-step grid. The follow exercises in table 11 shows the expected results:
Table 11
33
FORMULA = 6x - 36
43
44
: ( 6 x 53) - 36 = 282
53
54
55
2: 33+43+44+53+54+55=282
77
FORMULA = 6x - 36
87
88
: (6 x 97) - 36 = 546
97
98
99
2: 77+87+88+97+98+00=546
3
FORMULA = 6x - 40
24
25
: (6 x 35) - 36 = 170
35
36
37
2: 13+24+25+35+36+37=170
51
FORMULA = 6x - 40
62
63
: (6 x 73) - 40 = 398
73
74
75
2: 51+62+63+73+74+75=398
21
FORMULA = 6x - 44
33
34
: (6 x 45) - 44 = 226
45
46
47
2: 21+33+34+45+46+47=226
53
FORMULA = 6x - 44
65
66
: (6 x 77) - 44 = 418
77
78
79
2: 53+65+66+77+78+79=418
From the above 6 examples, it clearly indicates that the algebra equation works and the theory is accurate.
In this exercise the method of adding the numbers was used to ensure that the algebra equation gives the correct results and in all cases it did, proving the equation to be correct.
From the above exercises and the results we can state the hypothesis as:
. Each time the step stair increases so does the amount of squares within the grid, which is called the formula of triangular numbers.
2. Also as the size of the grid increase by x number of squares the value of the squares total will also change again known as the triangular numbers.
We continue with the exercise and so far we have our algebra equation is 6 x - 36 for a 3-step grid in a 10x10 boxed grid
6 x - 40 for an 11x11 boxed grid and 6 x - 44 for a 12x12 boxed grid.
We continue to proceed with the exercise to complete our formula (The General Formula). In the algebra equation we need ...
This is a preview of the whole essay
2. Also as the size of the grid increase by x number of squares the value of the squares total will also change again known as the triangular numbers.
We continue with the exercise and so far we have our algebra equation is 6 x - 36 for a 3-step grid in a 10x10 boxed grid
6 x - 40 for an 11x11 boxed grid and 6 x - 44 for a 12x12 boxed grid.
We continue to proceed with the exercise to complete our formula (The General Formula). In the algebra equation we need to include the value of the grid size, i.e. is it a 4x4 [4] or 16x16 [16] represented as algebra to give us the total of the numbers added together.
To start the exercise we need to establish the highest common factor, by using our values above, i.e. 36,40 & 44 and show them as shown below:
We know that the above values increase by the constant number [4] and also that in any 3-step grid square if the grid size increases by 1 then the constant number is added to the value (The highest Common Factor)
In the above illustration the grid size is increasing by 1, i.e. from 10 to 11 and then to 12, using the highest common factor for 3-step grids (4) the calculations are 9, 10 & 11 i.e:
9 x 4 = 36 10 x 4 = 40 11 x 4 = 44
We can now use 4 as the constant number and [n] as the grid size in our algebra equation. Therefore the algebra equation starts to look like this:
6 x - 4n The next step is to test the equation, we will use 15 as the grid size.
n = 15 and our 3-step grid numbers are 1,16,17,31,32,33 as shown below and x = 31:
Table 12:
6
7
31
32
33
Using the equation 6 x - 4(n) we can use the above values to see if the equation works
(6 x 31) - (4 x 15) = 126
To check the equation is correct we added all the numbers:
+16+17+31+32+33 = 130 Therefore 126 <> 130 and our equation is incorrect.
The value from the equation is higher that the correct result, therefore we need to reduce n, and we can start this by 1 i.e. (n - 1).
For example 6 x - 4 (n - 1)
We can again test this using the above 3-step grid.
(6 x 31) - (4 x (15 - 1)
86 - 56 = 130
30 = 130 therefore the result from our equation is the same as it is by adding up the numbers.
To prove our algebra equation is correct we test it on two other 3-step grids, 32 & 45 as shown below:
Table 13 a 32x32 and 45x45 grid:
33
34
46
47
65
66
67
91
92
93
Using our equation we use the [x] and [n] values we see the results below:
(6 x 65) - (4 x (32 - 1) = 266 and also 1+33+34+65+66+67 = 266
Add for the 45x45 grid box the result is:
(6 x 91) - (4 x (45 - 1) = 370 and also 1+46+47+91+92+93 = 370
From the steps we have taken we have formulated an algebra equation which when tested, gives us positive results in every case for a 3-step stair in any grid size:
THE GENERAL FORMULA IS:
Assignment Part 2
Part 2 Objective
Investigate further the relationship between the stair totals and other step stairs on other number grids.
Using our algebra equations and our proven theory for 4-step, 5-step and 6-step stairs for any numbered grid size. using the same approach and theory as we used in the 3-step exercise we can apply the same to the 4-step stair to find the general formula
Four step stairs:
0 x 10 Grid
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
00
x-30
x-20
x-19
x-10
x-9
x-8
X
X+1
X+2
X+3
For example if x = 31 then the square above would be x - 10 = 11 (31 - 10 = 21) and so on, Using the algebra equation and adding the values we get is -90
1 x 11 Grid
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
00
01
02
03
04
05
06
07
08
09
10
x-33
x-22
x-21
x-11
x-10
x-9
X
X+1
X+2
X+3
For example if x = 34 then the square above would be x - 11 = 12 (34 - 11 = 23) and so on, Using the algebra equation and adding the values we get -100
2 x 12 Grid
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
00
01
02
03
04
05
06
07
08
09
10
11
12
13
14
15
16
17
18
19
20
x-36
x-24
x-23
x-12
x-11
x-10
x
X+1
X+2
X+3
For example if x = 37 then the square above would be x - 12 = 25 (37 - 12 = 25) and so on Using the algebra equation and adding the values we get -110
If we closely examine the results from the 4-step stairs from the 3 numbered Grid Square, i.e. 10x10, 11x11 and 12x12 we can see that the constant value [10]
We can then complete the rest of the table by using these numbers square as shown below:
From the table below and using our results from the investigation we can see a pattern emerging. Every time the size of the grid square increases, the value in the algebra formula increases by 10. For example for a 10x10 numbered grid using a 4-step stair the formula is 10x-90 then we increase the grid size by 1, 11x11 and using the same 4-stepped stair approach the formula is 10x-100, etc. We can clearly see the constant number [10] is consistent every time the grid size increases by [1].
For any 4-step numbered grid box as shown below, using the bottom left grid box's value as x and the algebra theory used to calculate the equation the value of the box can be found. This theory has been put to the test using a 10x10, 11x11 and 12x12 grid boxes. The results are conclusive and consistent, proving the theory to be accurate and reliable.
Using any of the algebra equation from the above table, e.g. 10x-90 or 10x-110 we can prove the results for any 4-step grid:
21
Formula: 10x- 90
31
32
: 10 x 51 - 90= 420
41
42
43
2: 21+31+41+51+32+42+52+43+53+54= 420
51
52
53
54
37
Formula: 10x- 90
47
48
: 10 x 67 - 90= 580
57
58
59
2: 37+47+57+67+48+58+68+59+69+70= 580
67
68
69
70
24
Formula: 10x- 100
35
36
: 10 x - 100= 470
46
47
48
2: 24+35+46+57+36+47+58+48+59+60= 470
57
58
59
60
51
Formula: 10x- 100
62
63
: 10 x - 100= 740
73
74
75
2: 51+62+73+84+63+74+85+75+86+87= 740
84
85
86
87
30
Formula: 10x- 110
42
43
: 10 x - 110= 740
54
55
56
2: 30+42+54+66+43+55+67+56+68+69= 740
66
67
68
69
From the above 5 examples of such exercise, it clearly indicates that the algebra equation works and the theory is accurate.
In this exercise the method of adding the numbers was used to ensure that the algebra equation gives the correct results and in all cases it did, proving our equation to be correct.
We continue to proceed with the exercise to complete our formula (The General Formula). In the algebra equation we need to include the value of the grid size, i.e. is it a 4x4 [4] or 16x16 [16] represented as algebra to give us the total of the numbers added together.
To start the exercise we need to establish the highest common factor, by using our values above, i.e. 90,100 & 100 and show them as shown below:
We know that the above values increase by the constant number [10] and also that in any 4-step grid square if the grid size increases by 1 then the constant number is added to the value (The highest Common Factor)
In the above illustration the grid size is increasing by 1, i.e. from 10 to 11 and then to 12, using the highest common factor for 4-step grids (10) the calculations are 9, 10 & 11 i.e:
9 x 10 = 90 10 x 10 = 100 11 x 10 =110
We can now use 10 as the constant number and [n] as the grid size in our algebra equation. Therefore the algebra equation starts to look like this:
0 x - 10(n) and the next step is to test the equation; we will use 15 as the grid size.
n = 15 and our 4-step grid numbers are 1,16,17,31,32,33,46,47,48,49 as shown below and x = 46:
6
7
31
32
33
46
47
48
49
Using the equation 10 x - 10(n) we can use the above values to see if the equation works
(10 x 46) - (4 x 15) = 400 To check the equation is correct we added all the numbers:
+16+17+31+32+33+46+47+48+49 = 320 Therefore 400 <> 320 and our equation is incorrect.
The value from the equation is lower that the correct result, therefore we need to reduce n, and we can start this by 1 i.e. (n - 1).
For example 10 x - 10 (n - 1) we can again test this using the above 3-step grid.
(10 x 46) - (10 x (15 - 1) 460 - 140 = 320
320 = 320 therefore the result from our equation is the same as it is by adding up the numbers.
From the steps we have taken we have formulated an algebra equation which when tested, gives us positive results in every case:
THE GENERAL FORMULA IS:
Five step stairs: 10 x 10 Grid
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
00
x-40
x-30
x-29
x-20
x-19
x-18
x-10
x-9
x-8
x-7
X
X+1
X+2
X+3
X+4
Using the algebra equation and adding the values we get -180
1 x 11 Grid
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
00
01
02
03
04
05
06
07
08
09
10
x-44
x-33
x-32
x-22
x-21
x-20
x-11
x-10
x-9
x-8
X
X+1
X+2
X+3
X+4
Using the algebra equation and adding the values we get -200
2 x 12 Grid
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
00
01
02
03
04
05
06
07
08
09
10
11
12
13
14
15
16
17
18
19
20
x-48
x-36
x-35
x-24
x-23
x-22
x-12
x-11
x-10
x-9
X
X+1
X+2
X+3
X+4
Using the algebra equation and adding the values we get -220
If we closely examine the results from the 5-step stairs from the 3 numbered grid boxes, i.e. 10x10, 11x11 and 12x12 we can see there is a constant number that is consistent, which is [20]
We can then complete the rest of the table by using this constant number as shown below:
From the table below and using our results from the investigation we can see a pattern emerging. Every time the size of the grid square increases, the value in the algebra formula increases by 20. For example for a 10x10 numbered grid using a 5-step stair the formula is 15x-180 then we increase the grid size by 1, 11x11 and using the same 5-stepped stair approach the formula is 15x-200, etc. We can clearly see the constant number [20] is consistent every time the grid size increases by [1].
For any 5-step numbered grid box as illustrated, using the bottom left grid box's value as x and the algebra theory used to calculate the equation the value of the box can be found. This theory has been put to the test using a 10x10, 11x11 and 12x12 grid boxes. The results are conclusive and consistent, proving the theory to be accurate and reliable.
Using any of the algebra equation from the above table, e.g. 15x-180 or 15x-220 we can prove the results for any 5-step grid:
1
Formula: 15x-180
21
22
: 15x51-180= 585
31
32
33
2: 21+31+41+51+32+42+52+43+53+54+11+22+33+44+55= 585
41
42
43
44
51
52
53
54
55
3
Formula: 15x-180
23
24
: 15x53-180= 615
33
34
35
2: 13+23+33+43+53+24+34+44+54+35+45+55+46+56+57= 615
43
44
45
46
53
54
55
56
57
3
Formula: 15x-200
24
25
: 15x57-200= 655
35
36
37
2: 24+35+46+57+36+47+58+48+59+60+13+25+37+49+61= 655
46
47
48
49
57
58
59
60
61
2
Formula: 15x-200
23
24
: 15x56-200= 640
34
35
36
2: 12+23+34+45+56+24+35+46+57+36+47+58+48+59+60= 640
45
46
47
48
56
57
58
59
60
8
Formula: 15x-220
30
31
: 15x66-220= 770
42
43
44
2: 30+42+54+66+43+55+67+56+68+69+18+31+44+57+70= 770
54
55
56
57
66
67
68
69
70
Formula: 15x-220
3
4
: 15x49-220= 515
25
26
27
2: 1+13+25+37+49+14+26+38+50+27+39+51+40+52+53= 515
37
38
39
40
49
50
51
52
53
From the above 5 examples of such exercise, it clearly indicates that the algebra equation works and the theory is accurate.
In this exercise the method of adding the numbers was used to ensure that the algebra equation gives the correct results and in all cases it did, proving our equation to be correct.
We continue to proceed with the exercise to complete our formula (The General Formula). In the algebra equation we need to include the value of the grid size, i.e. is it a 4x4 [4] or 16x16 [16] represented as algebra to give us the total of the numbers added together.
To start the exercise we need to establish the highest common factor, by using our values above, i.e. 280,200 & 220 and show them as shown below:
We know that the above values increase by the constant number [20] and also that in any 5-step grid square if the grid size increases by 1 then the constant number is added to the value (The highest Common Factor)
In the above illustration the grid size is increasing by 1, i.e. from 10 to 11 and then to 12, using the highest common factor for 5-step grids (20) the calculations are 9, 10 & 11 i.e:
9 x 20 = 180 10 x 20 = 200 11 x 20 =220
We can now use 20 as the constant number and [n] as the grid size in our algebra equation. Therefore the algebra equation starts to look like this:
5 x - 20(n) The next step is to test the equation; we will use 15 as the grid size.
n = 15 and our 5-step grid numbers are 1,16,17,31,32,33,46,47,48,49,61,62,63,64,65 as shown below and x =61:
6
7
31
32
33
46
47
48
49
61
62
63
64
65
Using the equation 15 x - 20(n) we can use the above values to see if the equation works
(15 x 61) - (20 x 15) = 615 To check the equation is correct we added all the numbers:
+16+17+31+32+33+46+47+48+49+61+62+63+64+65 = 635
Therefore 615 <> 635 and our equation is incorrect.
The value from the equation is lower that the correct result, therefore we need to reduce n, and we can start this by 1 i.e. (n - 1).
For example 15 x - 20 (n - 1) We can again test this using the above 5-step grid.
(15 x 61) - (20 x (15 - 1) 915 - 280 = 635
635 = 635 therefore the result from our equation is the same as it is by adding up the numbers.
From the steps we have taken we have formulated an algebra equation which when tested, gives us positive results in every case:
THE GENERAL FORMULA IS:
Six step stairs:
0 x 10 Grid
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
00
X-50
X-40
X-39
X-30
X-29
X-28
X-20
X-19
X-18
X-17
X-10
X-9
X-8
X-7
X-6
X
X+1
X+2
X+3
X+4
X+5
Using the algebra equation and adding the values we get -315
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
00
01
02
03
04
05
06
07
08
09
10
X-55
X-44
X-43
X-33
X-32
X-31
X-22
X-21
X-20
X-19
X-11
X-10
X-9
X-8
X-7
X
X+1
X+2
X+3
X+4
X+5
Using the algebra equation and adding the values we get -350
2 x 12 Grid
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
00
01
02
03
04
05
06
07
08
09
10
11
12
13
14
15
16
17
18
19
20
X-60
X-48
X-47
X-36
X-35
X-34
X-24
X-23
X-22
X-21
X-12
X-11
X-10
X-9
X-8
X
X+1
X+2
X+3
X+4
X+5
Using the algebra equation and adding the values we get -385
If we closely examine the results from the 6-step stairs from the 3 numbered grid boxes, i.e. 10x10, 11x11 and 12x12 we can see there is a constant number that is consistent, which is [35]
We can then complete the rest of the table by using this incremental number as shown below:
From the table below and using our results from the investigation we can see a pattern emerging. Every time the size of the grid square increases, the value in the algebra formula increases by 35. For example for a 10x10 numbered grid using a 6-step stair the formula is 21x-315 then we increase the grid size by 1, 11x11 and using the same 6-stepped stair approach the formula is 21x-350, etc. We can clearly see the constant number [35] is consistent every time the grid size increases by [1].
For any 6-step numbered grid box as shown below, using the bottom left grid box's value as x and the algebra theory used to calculate the equation the value of the box can be found. This theory has been put to the test using a 10x10, 11x11 and 12x12 grid boxes. The results are conclusive and consistent, proving the theory to be accurate and reliable.
Using any of the algebra equation from the above table, e.g. 21x-315 or 21x-385 we can prove the results for any 6-step grid:
Formula: 21x-315
1
2
: 21x51-315= 756
21
22
23
+11+21+31+41+51+12+22+32+42+52+23+33+43+
2: 53+34+44+54+45+55+56= 756
31
32
33
34
41
42
43
44
45
51
52
53
54
55
56
3
Formula: 21x-315
23
24
: 21x63-315= 1008
33
34
35
2: 13+23+33+43+53+63+24+34+44+54+64+35+45+
55+65+46+56+66+57+67+68= 1008
43
44
45
46
53
54
55
56
57
63
64
65
66
67
68
24
Formula: 21x-350
35
36
: 21x79-350= 1309
46
47
48
2: 24+35+46+57+68+79+36+47+58+69+70+81+80+
48+59+60+71+82+72+83+84= 1309
57
58
59
60
68
69
70
71
72
79
80
81
82
83
84
37
Formula: 21x-350
48
49
: 21x92-350= 1582
59
60
61
2: 37+48+59+70+81+92+49+60+71+82+93+61+72+
83+94+73+84+95+85+96+97= 1582
70
71
72
73
81
82
83
84
85
92
93
94
95
96
97
Formula: 21x-385
3
4
: 21x61-385= 896
25
26
27
2: 1+13+25+14+26+27+37+38+39+40+49+50+51+
52+53+61+62+63+64+65+66= 896
37
38
39
40
49
50
51
52
53
61
62
63
64
65
66
40
Formula: 21x-385
52
53
: 21x100-385= 1715
64
65
66
2: 40+52+64+76+88+100+53+65+77+89+101+66+
78+90+102+79+91+103+92+104+105= 1715
76
77
78
79
88
89
90
91
92
00
01
02
03
04
05
From the above 5 examples of such exercise, it clearly indicates that the algebra equation works and the theory is accurate.
In this exercise the method of adding the numbers was used to ensure that the algebra equation gives the correct results and in all cases it did, proving our equation to be correct.
We continue to proceed with the exercise to complete our formula (The General Formula). In the algebra equation we need to include the value of the grid size, i.e. is it a 4x4 [4] or 16x16 [16] represented as algebra to give us the total of the numbers added together.
To start the exercise we need to establish the highest common factor, by using our values above, i.e. 315,350 & 385 and show them as shown below:
We know that the above values increase by the constant number [35] and also that in any 6-step grid square if the grid size increases by 1 then the constant number is added to the value (The highest Common Factor)
In the above illustration the grid size is increasing by 1, i.e. from 10 to 11 and then to 12, using the highest common factor for 6-step grids (35) the calculations are 9, 10 & 11 i.e:
9 x 35 = 315 10 x 35 = 350 11 x 35 =385
We can now use 35 as the constant number and [n] as the grid size in our algebra equation. Therefore the algebra equation starts to look like this:
21 x - 35(n) the next step is to test the equation; we will use 15 as the grid size.
n = 15 and our 5-step grid numbers are 1,16,17,31,32,33,46,47,48,49,61,62,63,64,65,76,77,78,79,80,81 as shown below and x =76:
6
7
31
32
33
46
47
48
49
61
62
63
64
65
76
77
78
79
80
81
Using the equation 21 x - 35(n) we can use the above values to see if the equation works
(21 x 76) - (35 x 15) = 1071 To check the equation is correct we added all the numbers:
+16+17+31+32+33+46+47+48+49+61+62+63+64+65+76+77+78+79+80+81 = 1106 Therefore 1071 <> 1106 and our equation is incorrect.
The value from the equation is lower that the correct result, therefore we need to reduce n, and we can start this by 1 i.e. (n - 1).
For example 21 x - 35 (n - 1) We can again test this using the above 6-step grid.
(21 x 76) - (35 x (15 - 1) 1596 - 490 = 1106
106 = 1106 therefore the result from our equation is the same as it is by adding up the numbers.
From the steps we have taken we have formulated an algebra equation which when tested, gives us positive results in every case:
THE GENERAL FORMULA IS:
Conclusion
From the investigation and the exercises conducted in this assignment we can conclude that by using the general formula to calculate the total of the grid numbers:
* The step stairs, regardless to whether it is a 3-stepped, 4-stepped, or 7-stepped etc or where its position is in a numbered grid, i.e. 10x10, or 21x21 etc the general formula will always give the total of the gird numbers in that step stair.
* The investigations carried out in part 1 and part 2 show clearly illustrates the common relationship when using the general formula
The Universal Formula
The aim of a Universal Formula is to calculate the total value of the numbers in a step stair.
In the hypothesis we stated the notion of Triangular Numbers, whereby when there is an increase in the number of steps in a numbered grid the total of the squares increase.
Below is a representation of triangular numbers:
P = 1
3n
P = 2
6n
P = 3
0n
P = 4
5n
P = 5
21n
From the triangular numbers diagram the first set of numbers are referred to as the "a" numbers (1,4,10,20,35,56), 2nd set as the "c" numbers (1st difference - 3,6,10,15,21,28), 3rd set as the "d" numbers (2nd difference - 3,4,5,6,7) and finally the last set "e" numbers (common difference all the 1's). p is referred to as the step index.
Therefore if p = 5 then there will be 21 squares (1st difference) in the step stair, making it a 6-step stair as shown on the right:
Example using a 16x16 grid
A mathematical formula has already been written, referred to as the "Universal Formula" as shown below:
[1/2 p2 + (3p/2) + 1] X (x) - [1 + (p - 1)/6 (p2 + 4p + 6)] X (n - 1)
Where p is the number in a triangular grid, x = the value of the bottom left grid in the step stair and n = the size of the grid
Taking our 16x16 grid example we can calculate the total number of the value for the step stair:
P = 5 x = 133 n = 16
[[1/2 (5 x 5) + ((3 x 5)/2) + 1] X (x)] = 21x -
[1 + ((5 - 1) ((5 x 5) + (4 x 5) + 6))/6 x (n - 1) = 35 (n - 1)
= 21x - 35 (n - 1) = (21 x 133) - (35 x 15)
= 2793 - 525 = 2268
To test the answer we add the numbers:
53+69+70+85+86+87+101+102+103+104+117+118+119+120+121+133+134+135+136+137+138 = 2268
This demonstrates that the universal formula works for any step stair, any numbered grid size and combination of numbers:
We can do one more test using the same step stair, but starting the step at 104 as shown on the side
Therefore: p = 2 because there are 6 squares i.e. 6n and x = 136 and n = 16
Using the universal formula we can calculate:
[[1/2 (2 x 2) + ((3 x 2)/2) + 1] X (x)] = 6x - [1 + ((2 - 1) ((2 x 2) + (4 x 2) + 6))/6 x (n-1)=4(n - 1)
= 6x - 4 (n - 1) = (6 x 136) - (4 x 15)
= 816 - 60 = 756 Add the numbers: 104+120+121+136+137+138 = 756
The formula shows that no matter what the size of the step stair (i.e. number of squares), the position of the step stair (e.g. 1,11,12,21,22,23 or 81,82,83,91,92,93) and the grid size (e.g. 10x10, 25x25) the result will always be the total of the numbers in the step stair.
This relationship is known as the "triangular numbers"
GCSE Mathematics Course work Tejesh Patel, Class 11h
Page 21
GCSE Mathematics Course work Tejesh Patel, Class 11h
Page 2