At the T-Number the common difference is 10.
For example take 20 and 30 subtract them (30 -20) and you get 10.
Therefore the formula of this is 10n ± C = T-Number/ n ± C = 20, 20 - 10=10.
Therefore nth term= 10n+10=20
At the T-Total, the common difference is 50.
For example take 37 and 87 subtract them (87-37) and you get 50. Therefore the formula of this is 50n ± C = T-Total/ 5n ± C = 37, 50 - 37= 13.
Therefore nth term= 50n -13=37
Grid 1: 9 by 9 – Down ( )
Formula Expressed in the nth term
As the tables show every time the T-Number increases by 9 the T-Total increases by 45. I will express in the nth term by using these common differences and start on the basis of ‘if n=1’ but change accordingly if appropriate.
At the T-Number the common difference is 9.
For example take 20 and 29 subtract them (29 -20) and you get 9.
Therefore the formula of this is 9n ± C = T-Number/ 9n ± C = 20, 20-9=11.
Therefore nth term= 9n+11=20
At the T-Total, the common difference is 45.
For example take 37 and 82 subtract them (82-37) and you get 45. Therefore the formula of this is 45n ± C = T-Total/ 45n ± C = 37, 45-37=8.
Therefore nth term=45n-8=37.
Across- Grid 2: 5 by 5
Formula Expressed in the nth term
As the tables show every time the T-Number increases by 1 the T-Total increases by 5. I will express in the nth term by using these common differences and start on the basis of ‘if n=1’ but change accordingly if appropriate.
At the T-Number the common difference is 1.
For example take 17 and 18 subtract them (18 -17) and you get 1.
Therefore the formula of this is 1n ± C = T-Number/ n ± C = 17, 17 -1=16
Therefore nth term= n+16=17
At the T-Total, the common difference is 5.
For example take 75 and 80 subtract them (80-75) and you get 5. Therefore the formula of this is 5n ± C = T-Total/ 5n ± C = 75, (if n=17 (T-Number) ) 17 x 5= 85, 85-( 2times common difference)10= 75.
Therefore nth term= 5n-10=75
Across- Grid 3: 6 by 6
Formula Expressed in the nth term
As the tables show every time the T-Number increases by 1 the T-Total increases by 5. I will express in the nth term by using these common differences and start on the basis of ‘if n=1’ but change accordingly if appropriate.
At the T-Number, the common difference is 1.
For example take 14 and 15 subtract them (15 -14) and you get 1.
Therefore the formula of this is 1n ± C = T-Number/ n ± C = 14, 14 -1=13.
Therefore nth term= n+13=14
At the T-Total, the common difference is 5.
For example take 28 and 33 subtract them (33-28) and you get 5. Therefore the formula of this is 5n ± C = T-Total/ 5n ± C = 28, (if n=14, the T- Number) 14 x 5= 70 – 28=42
Therefore nth term=5n-42=28.
Across- Grid 4: 7 by 7
Formula Expressed in the nth term
As the tables show every time the T-Number increases by 1, the T-Total increases by 5. I will express in the nth term by using these common differences and start on the basis of ‘if n=1’ but change accordingly if appropriate.
At the T-Number, the common difference is 1.
For example take 16 and 17 subtract them (17 -16) and you get 1.
Therefore the formula of this is 1n ± C = T-Number/ n ± C = 16, 16 -1=15.
Therefore nth term= n+15=16.
At the T-Total, the common difference is 5.
For example take 21 and 36 subtract them (31-36) and you get 5. Therefore the formula of this is 5n ± C = T-Total/ 5n ± C = 31, (if n=16 (T-Number) ) 16 x 5= 80, 80 -16= 64
Therefore nth term= 5n-64=22
Across - Grid 5: 10 by 10
Formula Expressed in the nth term
As the tables show every time the T-Number increases by 1, the T-Total increases by 5. I will express in the nth term by using these common differences and start on the basis of ‘if n=1’ but change accordingly if appropriate.
At the T-Number, the common difference is 1.
For example take 22 and 23 subtract them (22 -23) and you get 1.
Therefore the formula of this is 1n ± C = T-Number/ n ± C = 22, 22 - 1=21
Therefore nth term= n+21=22
At the T-Total, the common difference is 5.
For example take 40 and 45 subtract them (45-40) and you get 5. Therefore the formula of this is 5n ± C = T-Total/ 5n ± C = 40, (if n=22 (T-Number) ) 22 x 5= 110, 110 - 40= 70.
Therefore nth term= 5n-70=40
Across - Grid 6: 11 by 11
Formula Expressed in the nth term
As the tables show every time the T-Number increases by 1, the T-Total increases by 5. I will express in the nth term by using these common differences and start on the basis of ‘if n=1’ but change accordingly if appropriate.
At the T-Number, the common difference is 1.
For example take 24 and 25 subtract them (24 -25) and you get 1.
Therefore the formula of this is 1n ± C = T-Number/ n ± C = 24, 24 - 1=23
Therefore nth term= n+23=24
At the T-Total, the common difference is 5.
For example take 41 and 46 subtract them (46-41) and you get 5. Therefore the formula of this is 5n ± C = T-Total/ 5n ± C = 41, (if n=24 (T-Number) ) 24 x 5= 120, 120 - 41= 79
Therefore nth term= 5n-79=41
Extension: Transformations Grid 7: 9 by 9 (Going across the grid)
180 degrees 90 degrees
270
degre-es
T at 180 degrees ( )
Formula Expressed in the nth term
As the tables show every time the T-Number increases by 1, the T-Total increases by 5. I will express in the nth term by using these common differences and start on the basis of ‘if n=1’ but change accordingly if appropriate.
At the T-Number, the common difference is 1.
For example take 10 and 11 subtract them (11 -10) and you get 1.
Therefore the formula of this is 1n ± C = T-Number/ n ± C = 10, 10 - 1=9
Therefore nth term= n+9=10
At the T-Total, the common difference is 5.
For example take 57 and 62 subtract them (62-57) and you get 5. Therefore the formula of this is 5n ± C = T-Total/ 5n ± C = 57, (if n=10 (T-Number) ) 10 x 5= 50, 57 - 50= 7
Therefore nth term= 5n+7=57
T at 90 degrees ( )
Formula Expressed in the nth term
As the tables show every time the T-Number increases by 1, the T-Total increases by 5. I will express in the nth term by using these common differences and start on the basis of ‘if n=1’ but change accordingly if appropriate.
At the T-Number, the common difference is 1.
For example take 2 and 3 subtract them (3 -2) and you get 1.
Therefore the formula of this is 1n ± C = T-Number/ n ± C = 2, 2 - 1=1
Therefore nth term= n+1=2
At the T-Total, the common difference is 5.
For example take 73 and 78 subtract them (78-73) and you get 5. Therefore the formula of this is 5n ± C = T-Total/ 5n ± C = 73, (if n=2 (T-Number) ) 2 x 5= 10, 10 - 73= 63
Therefore nth term= 5n+63=73
T at 270 degrees ( )
Formula Expressed in the nth term
As the tables show every time the T-Number increases by 1, the T-Total increases by 5. I will express in the nth term by using these common differences and start on the basis of ‘if n=1’ but change accordingly if appropriate.
At the T-Number, the common difference is 1.
For example take 12 and 13 subtract them (13 -12) and you get 1.
Therefore the formula of this is 1n ± C = T-Number/ n ± C = 12, 12 - 1=11
Therefore nth term= n+11=12
At the T-Total, the common difference is 5.
For example take 53 and 58 subtract them (58-53) and you get 5. Therefore the formula of this is 5n ± C = T-Total/ 5n ± C = 53, (if n=12 (T-Number) ) 12 x 5= 60, 60 - 53= 7
Therefore nth term= 5n-7=53
Conclusion
Going across the grid, I found that the Trend of T-Numbers was always +1 and the trend in T-Totals was always +5. This was even apparent in the transformations of the T shape. This trend caused the nth term for all of the T-Numbers to always include 1n/n and the nth term of T-Totals to always include 5n. As you can see from my results, the following nth terms going across the grid were:
- For a 9 by 9 grid – T-number= n+19=20, T-Total=5n-63=37
- For a 5 by 5 grid – T-number= n+16=17, T-Total=5n-10=75
- For a 6 by 6 grid – T-number= n+13=14, T-Total=5n-42=28
- For a 7 by 7 grid – T-number= n+15=16, T-Total=5n-64=22
- For a 10 by 10 grid – T-number= n+21=22, T-Total=5n-70=40
- For a 11 by 11 grid – T-number= n+23=24, T-Total=5n-79=41
- For a 180 degree transformation at a 9 by 9 grid - T-number=n+9=10, T-Total=5n+7=57
- For a 90 degree transformation at a 9 by 9 grid - T-number= n+1=2, T-Total=5n+63=73
- For a 270 degree transformation at a 9 by 9 grid – T-number=n+11=12, T-Total= 5n-7=53
As you can see this supports my prediction showing all T-Numbers include 1n/n and T-totals include 5n.
From my results, I can also predict that T-Shapes going down a grid would have a nth term including 9n for T-Numbers and 45n for T-totals. This is shown in my results by: the nth term for T-Numbers going down a 9 by 9 grid is 9n+11=20 and the nth term for a T-Total going down a 9 by 9 grid is 45n-8=37.
Also going diagonally down a grid I can predict that T-Shapes have a nth term including 10n for T-Numbers and 50n for T-Totals. This is shown by in my results by: the nth term for T-Numbers going diagonally down a 9 by 9 grid is 10n+10=20 and the nth term for a T-Total going down a 9 by 9 grid is 50n -13=37.