178
149+
327
N=22
T=3+4+12+21+22=
62
From above I can see a pattern forming; the T numbers are ending in a 2 or a 7:
252, 87, 327, and 62.
Now I am going to draw some more C’s but this time by translating hem sideways in a logical way.
MY PREDICTION
I predict that the T numbers are going to go up in 5’s as I slide the C’s sideways because when the T numbers: 7 and 2 are both taken away from each other: 7-2 =5
N=20
T=1+2+10+19+20=
52
N=21
T=2+3+11+20+21=
57
N=22
T=3+4+12+21+22=
62
N=23
T=4+5+13+22+23=
67
N=24
T=5+6+14+23+24=
72
N=25
T=6+7+15+24+25=
77
A TABLE SHOWING MY RESULTS…
+5 +5 +5 +5 +5
I can see that my prediction was right because all the numbers go up in 5’s.
THE FORMULA…
I have already established from the table that the formula will have 5 in it somewhere, this is because the numbers as shown in the table go up in 5’s.
The start of my formula is going to be like this:
T=5n-?? ~(?? =The 0th term)
Now I have to find the 0th term.
To do this I am going to multiply 20 by 5 giving us 100 then I am going to subtract 52-the 20th term from 100 giving me the 0th term, which is: -48
So now, I can complete my formula:
T=5n-48
EXPERIMENTATION WITH C’S…
Now I am going to translate the C downwards.
N=20
T=1+2+10+19+20=
52
Translating the C downward by one square…
N=29
T=10+11+19+28+29=
97
All the numbers seem to be in relation with the number 9:
~97-52=45
45\5=9
~RED=
10-1=9
~YELLOW=
11-2=9
~GREEN=
19-10=9
~BLUE=
28-19=9
~PINK=
29-20=9
I think that the number 9 keeps appearing is because the grid, which, I have drawn, is a 9x9 grid.
Below is a C like the other one’s but with n terms in it instead of the numbers from the grid.
From this I discovered that wherever in the 9x9 grid you go the N terms will always be the same. Though I do not think the C can be rotated etc.
Below I am going to draw a 9x9 grid again but this time I am going to alter the way I draw the C’s. This time I am going to put the C on its side like shown below:
So now, I am going to draw my 9x9 grid…
N=3
T=10+1+11+12+3=
37
N=4
T=11+2+12+13+4=
42
N=5
T=12+3+13+14+5=
47
I can already see a pattern forming; this pattern is similar to the one I discovered before. The numbers are going up in 5’s.
Now I am going to make a table showing my results.
I am going to now draw an example C with algebraic terms in it:
At this point, I am going to find the formula of C’s on their side.
5*5=25
47-25=22
This tells me the formula is:
T=5n+22
Moreover, if I check it, it works:
N=3
3*5=15+22
=T=37
EXPERIMENTING WITH GRIDS OF DIFFERENT SIZES…
Below I am going to draw a 4x4 grid, in which, like the 9x9 grid I am going to experiment with.
Now I am going to take some C’s and experiment with them
N=10
T=1+2+5+9+10=
27
N=11
T=2+3+6+10+11=
32
N=12
T=3+4+7+11+12=
37
N=14
T=5+6+9+13+14=
47
Because the grid is too small for me to see what the T number is going to be when the N number is 13. However, I assume it is going to be 42.
A table with my results…
The Formula is going to have the number 5 in it, because the T numbers go up in 5’s:
T=5n????
10*5=50
27-50=-23
Therefore the formula is:
T=5n-23
To check this:
N=10
5*10=50-23
=T=27
N=10
T=10+9+1+2+5=
27
N=14
T=5+6+9+13+14=
47
All the numbers, like the 9x9 grid, seem to be associated with the number 4.
~47-27=20
20/5=4
~ 5-1=4
~ 6-2=4
~ 9-5=4
~ 13-9=4
~ 14-10=4
I think that the number 4 keeps appearing is because the grid, which, I have drawn, is a 4x4 grid.
Mehjabeen Iqbal 10NDM Page