(x2 + 12x + 10) – (x2 + 12x)
= x2 + 12x + 10 – x2 – 12x
= 10
Thusly I have used algebra to prove that the difference is 10 and stays 10 wherever I place a 2 by 2 square on the grid. Now to test with a 3 by 3 square and see if it has a constant difference as well and if it is similar towards the 2 by 2 square.
1 x 23 = 23
3 x 21 = 63
63 – 23 = 40
The difference has come to 40 this time. To test if this is constant I will place the square in another place on the grid and analyse. I will place it randomly on the grid by using the number generator method and see if the results differ.
3 x 25 = 75
5 x 23 = 115
115 – 75 = 40
The difference is 40 again therefore this is constant and will as the 2 by 2 difference stay the same wherever I place it on the grid but to be 100% sure I need to use algebra to prove my theory.
x(x + 22) = x2 + 22x
(x + 2)(x + 20) = x2 + 12x + 40
(x2 + 12x + 40) – (x2 + 22x) = 40
I have used algebra to prove that the difference is 40.This is interesting as the fist difference is 10 and the next is 40 there possibly could be a pattern emerging here. I predict that the 4 by 4 square will have a difference of 70.I make this prediction because the first two differences are 30 apart from each other and the 4 by 4 square will be 40 +30 also. I might be a little presumptuous predicting the next difference along with only 2 results so far but lets see and find out if I am right.
I will now test the 4 by 4 theory and find the difference.
1 x 34 = 34
4 x 31 = 124
The difference is 90 this can’t be right because I had predicted 70 I will try again by randomly placing the square in another location.
5 x 38 = 190
8 x 35 = 280
280 – 190 = 90
Difference is 90 yet again, this is no mistake and my prediction is proving wrong, to test it one final time I will use algebra.
x(x + 33) = x2 + 33x
(x + 3)(x + 30) = x2 + 33x + 90
(x2 + 33x + 90) – (x2 + 33x) = 90
The difference is 90 and I have proven this with the use of algebra and numbers therefore my prediction was wrong and now I shall collate the differences so far and see what I can conclude on.
Here are my results so far:
This is interesting as the first difference isn’t constant but the second difference is 20.Looking at the sequence I predict that the 20 will remain the same for the second difference and the main difference for the 5 by 5 square will be 160.This is because I will add the 20 to the already increasing first difference of 50 which will be 70.Then I will add the 70 to the 4 by 4 main difference of 90 which will make 160.Therefore a 5 by 5 square in my opinion should have a difference of 160.Now I will test my new prediction.
1 x 45 = 45
5 x 41 = 205
205 – 45 = 160
x(x + 44) = x2 + 44x
(x + 4)(x + 40) = x2 + 44x + 160
(x2 + 44x + 160) – (x2 + 44x) = 160
My theory is correct and the difference has been proven to be 160.I have now found a pattern and have enough results to come up with a formula that will give me the results for any size square on a 10 by 10 size grid. I should be able to use this formula easily and at ease by cutting out all the working as I have done above for each square. Here is my table of results:
I have now predicted the rest of the results as I did with the 5 by 5 square. I can do this because I know the second difference will remain the same i.e. 20.
A graph can be shown to represent the table and how the different totals change. Looking at the difference it is increasing rapidly which is true as it is growing with an ever increasing first difference. As you can see the first difference is at a slant. It is increasing but what’s different is that it has a stable gradient that does not change therefore it is linear. Looking at the main difference it has a curved shape suggesting the equation of the line may be quadratic or squared equation as it resembles y = x2 .
This is where we can make formulae from the results of the table. The formula should connect the size of the square (n) which will be the length of one side of the square to the difference and save the hassle of the working out. Looking at the second difference which is 20 I know from my past number pattern work that when you end up with a constant second difference to start creating a formulae you must do a few things. Firstly half the number which will make the 20 into 10.Then since it is the second difference which is regular; the equation must include n squared. This sustains my prediction from the graph that the formula to work out the difference must include n squared as it has the distinct shape to it. Here is what we have so far:
Notice that the answer needed is the previous nth terms answer for instance when 3 is he n term the answer needed is 40, this is also the previous number 2’s outcome. This trend continues on with all the results. Therefore because n gives us an answer too big we can replace n with (n – 1) which will give me the correct answer. Therefore the formula in theory is,
10(n – 1) 2
Notice how the coefficient of n is 10 which is also the size of the grid, maybe this will be the same for an 11 by 11 grid. The formula also contains a square power because n is two terms not just one it is used as the same term twice.
Now I need to test this with a number in the table if I substitute 9 as n:
10(9 – 1) 2 = 640
I have the right answer therefore I have found a formula for finding any size square’s difference on a 10 by 10 grid. This might seem a lot but considering there are only 10 squares that can actually fit inside a 10 by 10 grid this is not a lot. The next logical step is to change the size of the grid and see how the results vary and if I can link the formula back to the one of the 10 by 10 grid. I will now test all the squares again on the 11 by 11 grid.
Firstly a 2 by 2 square I will find the difference. I will randomly place the 2 by 2 square on the grid then find the results.
3 x 13 = 39
2 x 14 = 28
Difference is 11
The algebra to back up the numbers:
x(x + 12) = x2 + 12x
(x + 1)(x + 11) = x2 + 12x + 11
Difference is 11
3 by 3:
5 x 25 = 125
3 x 27 = 81
Difference is 44
(x + 2)(x + 22) = x2 + 24x + 44
x(x + 24) = x2 + 24x
Difference is 44
4 by 4:
4 x 34 = 136
1 x 37 = 37
Difference is 99
(x + 3)(x + 33) = x2 + 36x + 44
x(x + 36) = x2 + 36x
Difference is 99
5 by 5:
5 x 45 = 225
1 x 49 = 49
Difference is 176
(x + 4)(x + 44) = x2 + 48x + 176
x(x + 48) = x2 + 48x
Difference is 176
I am starting to see a pattern with the 11 by 11 grid results here they are as follows:
From the first five results I have been able to predict the rest, this is because interestingly enough the second difference has remained the same for the first five and does not change. From this second difference I have been able to calculate the first and final difference. I know I have filled this table in correctly from what I have learnt about the 10 by 10 grid which also has a reoccurring second difference.
As with the 10 by 10 grid the difference line has the squared or quadratic shape to it. There fore the formula will be very similar to last one. We can already work out the first part of the formula because we have a recurring second difference. With my past work on number patterns I know that if the second difference is 22 there will be 11n 2 in the formula.
Here is what I have so far:
Notice again how the answer needed is always the previous nth terms outcome. Therefore the formula will be more or less the same as the 10 by 10 grid as it will contain (n – 1).In conclusion the formula I have worked out for any square on an 11 by 11 grid is
11(n - 1) 2
To test this I will take the number 8 from the table and see if I get the right difference. By substituting 8 into the formula I have got the following results:
11(8 – 1) 2
11 x 49 = 539
The result is correct and I have proven that the formula works. I now have 2 formulae and need to comment on them both and conclude on why they have certain characteristics to understand fully why they are the way they are. The formulae are as follows:
10 by 10 grid is 10(n - 1) 2
11 by 11 grid is 11(n – 1) 2
The formulae both have the size of the grid as the coefficient of the equation. I believe this is because the second difference is always the length or width of the grid doubled. Also both are to the power of 2 because the n term is being used to express a length and a width therefore it has to be squared. This is what leads me onto the next step of my investigation and that is investigating rectangular shapes where there are 2 variables and not just one. Although before I do this I can add to my formula. I can do this because by knowing that the coefficient of n is the size of the grid I can expand my formula. Therefore my formula now is:
x(n – 1) 2
x = length/width of grid
n= one side of the square
I have expanded my formula because now it will give me the difference from any size square within any size grid. To test my formula I will choose a 3 by 3 square on a 12 by 12 firstly work it out with the formula then the full way and see if the answers match.
With the formula:
x(n – 1) 2
12(3 – 1)2
The difference in theory is 48
I will now test the difference to see if it remains the same:
3 x 25 = 75
1 x 27 = 27
75 – 27 = 48
The difference is 48
This proves that my formula will work with any grid and any size square within it at any place on the grid. I have now reached the limit of how squares within a 2 dimensional grid can be investigated. To add another variable into my work I am going to investigate rectangles. This will add a further variable because as with squares both the length and width are represented by 1 number whereas rectangles can have 2 completely different numbers. Therefore I am expecting to work out a much different formula from the one above with the inclusion of another variable. This formula should provide me with the following:
- Difference to any size square or rectangle on a size grid in any location on the grid.
I will work out the difference to a 2 by 3 rectangle, all the work from this point will remain on a 10 by 10 grid until stated otherwise.
1 x 13 = 13
3 x 11 = 33
Difference is 20
(x + 2)(x + 10) = x2 + 12x + 20
x(x + 12) = x2 + 12x
Difference is 20
2 x 4:
4 x 11 = 44
1 x 14 = 14
Difference is 30
