I am now going to test the general formula by doing the differences for all square boxes in a 10 x 10 column grid.
2 x 2 = 10(2–1)2=10
3 x 3 = 10(3–1)2=40
4 x 4 = 10(4–1)2=90
5 x 5 = 10(5–1)2=160
6 x 6 = 10(6–1)2=250
7 x 7 = 10(7–1)2=360
8 x 8 = 10(8–1)2=490
9 x 9 = 10(9–1)2=640
10x10 = 10(10-1)2=810
Part 2 (Square box – Different grid size)
I am going to find the diagonal difference of 2 types of column grids 5 x 5 & 7 x 7. From this I will try to find out if there are any correlations by placing a square box.
5 x 5 Column Grid
I am going to find the diagonal difference of any 2 x2 box in a 5 column grid. From this I will try to find out if there are any correlations.
2 x 2 Box-Numeric’s
By looking at the 2 x 2 boxes in a 5 column grid, I have found out that if you place a 2 x 2 box, any where in the grid, the difference was 5.
Predict
Therefore I predict that, if I place any 2 x 2 box in a 5 column grid, the difference will always be 5.
This estimate proves my aim, that if you place any 2 x2 box in a 5 column grid, the difference will always be 5
Difference in Algebra
Now I am going to show what the difference will be for any 2 x 2 box in a 5 column grid algebraically
I am now going to increase the size of my boxes to 3 x 3, 4 x 4 etc and going to work out the difference in algebraic forms.
3 x 3 Box
4 x 4 Box
I am going to display my results in a table, where I would try finding any patterns within the results I have.
5 x 5 Grid
Therefore I can predict that the difference for a 5 x 5 grid will be 80. Now I am going to prove this prediction algebraically.
5 x 5 box - Proving algebraically
This shows that I have proved the difference for a 5 x 5 box in a 5 column grid correctly.
Now to find out a general formula for the difference for any square box in a 5 column grid will be.
Therefore, the overall formula for any square box in a 5 column grid is 5(n–1)2
I am now going to test the general formula by doing the differences for all square boxes in a 5 x 5 column grid.
2 x 2 = 5(2–1)2=5
3 x 3 = 5(3–1)2=20
4 x 4 = 5(4–1)2=45
5 x 5 = 5(5–1)2=80
7 x 7 Column Grid
I am going to find the diagonal difference of any 2 x2 box in a 7 column grid. From this I will try to find out if there are any correlations.
2 x 2 Box-Numeric’s
By looking at the 2 x 2 boxes in a 7 column grid, I have found out that if you place a 2 x 2 box, any where in the grid, the difference is 7.
Predict
Therefore I predict that, if I place any 2 x 2 box in a 7 column grid, the difference will always be 7.
This estimate proves my aim, that if you place any 2 x2 box in a 7 column grid, the difference will always be 7
Difference in Algebra
Now I am going to show what the difference will be for any 2 x 2 box in a 7 column grid algebraically
2 x 2 Box
I am now going to increase the size of my boxes to 3 x 3, 4 x 4 etc and going to work out the difference in algebraic forms.
3 x 3 Box
4 x 4 Box
I am going to display my results in a table, where I would try finding any patterns within the results I have.
7 x7 Grid
Therefore I can predict that the difference for a 7 x 7 grid will be 252. Now I am going to prove this prediction algebraically.
7 x 7 box - Proving algebraically
This shows that I have proved the difference for a 7 x 7 box in a 7 column grid correctly.
Now to find out a general formula for the difference for any square box in a 7 column grid will be.
Therefore, the overall formula for any square box in a 7column grid is 7(n–1)2
I am now going to test the general formula by doing the differences for all square boxes in a 10 x 10 column grid.
2 x 2 = 7(2–1)2=7
3 x 3 = 7(3–1)2=28
4 x 4 = 7(4–1)2=63
5 x 5 = 7(5–1)2=112
6 x 6 = 7(6–1)2=175
7 x 7 = 7(7–1)2=252
The general formula for the difference for any square box in any column grid will be shown below
Proving overall formula algebraically
Using the results gained from the calculations I can now prove and explain the formula g(b-1)2 for any square box in any grid size. I am explain this algebraically
1) n [(b-1)g + n + (b-1)]
=ng(b-1) + n2 + n(b-1)
[n+(b-1)] [n + g(b-1)]
= n2 + n(b-1) + ng(b-1) + g(b-1)2
2) n2 + n(b-1) + ng(b-1) + g(b-1)2 - ng(b-1) + n2 + n(b-1)
=g(b-1)2
Therefore this overall equation “g(b-1)2” explains the difference, for any size square box in any size grid.
Part 3 (10 column grid – Rectangular boxes)
I am going to find the diagonal difference of any 3 x2 box in a 10 column grid. From this I will try to find out if there are any correlations
3 x 2 Box-Numeric’s
By looking at the 3 x 2 boxes in a 10 column grid, I have found out that if you place a 3 x 2 box, any where in the grid, the difference will be 20.
Predict
Therefore I predict that, if I place any 3 x 2 box in a 10 column grid the difference
will always be 20.
This estimate proves my aim, that if you place any 3 x2 box in a 10 column grid, the difference will always be 20
Difference in Algebra
Now I am going to show what the difference will be for any 3 x 2 box in a 10 column grid algebraically.
I am now going to increase the size of my boxes to 4 x 5, 6 x 8 etc and going to work out the difference in algebraic forms.
4 x 5 Box
6 x 8 Box
I am going to display my results in a table, where I would try finding any patterns within the results I have.
10 x 10 Grid
Therefore I can predict that the difference for a 5 x 7 grid will be 240. Now I am going to prove this prediction algebraically.
5 x 7 Proving algebraically
By proving the difference for a 5 x 7 box in a 10 column grid, I can predict that the difference for a 12 x 9 box will be (12 – 1) = 11 ‘multiply’ (9 – 1) = 8. Therefore, it will be 11 x 8 = 88 (x10) = 880
Now to find out a general formula for the difference for any rectangle box in a 10 column grid will be
In the grid size c x r, (c) represents column and (r) represents rows. Therefore in the formula 10(c-1)(r-1) the (c) will be the number of columns and the (r) will be the rows of the box.
Proving formula algebraically
Using the results gained from the calculations I can now prove the formula 10(c-1)(r-1) for any rectangular box in a 10x10 grid. I am explain this algebraically
1) n[n+10(r-1)+(c-1)]
=n2+10n(r-1)+n(c-1)
[n+(c-1)]x[n+10(r-1)]
=n2+10n(r-1)+n(c-1)+10(c-1)(r-1)
2) n2+10n(r-1)+n(c-1)+10(c-1)(r-1) - n2+10n(r-1)+n(c-1)
=10(c-1)(r-1)
I am now going to test this formula 10(c-1)(r-1) on rectangular boxes in a 10 column grid.
3 x 2 =10(3-1)(2-1)=20
4 x 5 =10(4-1)(5-1)=120
5 x 6 =10(5-1)(6-1)=200
6 x 3 =10(6-1)(3-1)=100
5 x 7 =10(5-1)(7-1)=240
8 x 5 =10(8-1)(5-1)=280
9 x 7 =10(9-1)(7-1)=480
Part4 (Different grids –Different rectangular boxes)
By analysing my formula which I got for the second part, I can identify, that the pattern that encounters is only the change in grid size.
Therefore, by using this method, I will use the formula 10(c-1)(r-1) which I gained from part 3, however instead of the grid size 10, I will use (g) to represent my random grid size.
I will now represent this in a table.
Proving overall formula algebraically
I am now going to explain how I got this formula algebraically.
1) n[n+g(r-1)+(c-1)]
=n2+gn(r-1)+n(c-1)
[n+(c-1)][n+g(r-1)]
=n2+gn(r-1)+n(c-1)+g(c-1)(r-1)
2) n2+gn(r-1)+n(c-1)+g(c-1)(r-1) - n2+gn(r-1)+n(c-1)
=g(c-1)(r-1)
Therefore this overall equation “g(c-1)(r-1)” explains the difference, for any size rectangular box in any size grid.
Conclusion - In this task I have successfully found out general formula for any square box in any size grid which was to be ‘g(b-1)2’ and the general formula any rectangle box in any size grid which was to be ‘g(c-1)(r-1)’. These formulas were proven algebraically above.
If I were to extend this task further, I would change the grid numbers i.e. rather than it going up in 1s, it would go in 2s or 3s.
