Difference is found by subtracting the two products (bold numbers)
N (N + 33) – (N + 3) (N + 30)
N2 +33N – N2 + 3N +30N + 30
N2 +22N – N2 + 33N + 30
Difference = 90
Conclusion: the difference for a 4x4 squared box is always 90
Difference is found by subtracting the two products (bold numbers)
N (N + 44) – (N + 4 (N + 40)
N2 +44N – N2 + 4N +40N + 40
N2 +44N – N2 + 44N + 40
Difference = 160
Conclusion: the difference for a 5x5 squared box is always 160
The product difference for the different size square show pattern in their sequence
2x2 3x3 4x4 5x5 6x6
10 40 90 160 250 1st difference
30 50 70 90 2nd difference
20 20 20 3rd difference
The 3rd number difference is constant. Consequently now I can predict the next term or in this case the box number difference using the sequences
10 40 90 160 250
30 50 70 90
20 20 20
The next term can be predicted by adding 20 to the last term of the second sequence, which 90. Then add 90 to the last term in the first sequence, in this case it’s 160. Therefore 90 + 160 = 250.
This method can be used to predict the difference instead of the arithmetic way. I used this method to find the difference for the following by 7, 8 by 8, 9 by 9 and 10 by 10.
2x2 3x3 4x4 5x5 6x6 7x7 8x8 9x9 10x10
10 40 90 160 250 360 490 640 810 1st difference
30 50 70 90 110 130 150 170 2nd difference
20 20 20 20 20 20 20 3rd difference
I was provided with a formula to find the Nth an2 + bn + c. and all that was lift for me to do was to use this formula to find the Nth of my sequences.
c = 0 10 40 90 160 250 360 490 640 810 1st difference
a + b = 10 30 50 70 90 110 130 150 170 2nd difference
2 a = 20 20 20 20 20 20 20 20 3rd difference
First the value of a is found
2a = 20
a = 20 ÷ 2
a = 10
Then the value of b is found
a + b = 10
10 + b = 10
b = 10-10
b = 0
From the sequence I can already concluded that the value of c = 0. Now I will put the values in the formula of finding the nth term. The formula to find the Nth of a quadratic sequence is an2 + bn + c
a = 10
b = 0
c = 0
10n2 + 0n + 0 Formula cancels down to 10n2
I will check if my formula works
10 x 22 = 40 wrong answer
Since my formula didn’t work. When I substituted 2 in n’s place. I decided to do experimentations with my sequence and formula. I found out a way to get the answer but it involved the use of invented numbers, which didn’t include the sequence.
Theory:
1 2 3 4 5 6 7 8 9
2x2 3x3 4x4 5x5 6x6 7x7 8x8 9x9 10x10
10 40 90 160 250 360 490 640 810 1st difference
30 50 70 90 110 130 150 170 2nd difference
20 20 20 20 20 20 20 3rd difference
I decided instead of putting 2 in n’s place to find the difference for 2 by 2 boxes. I will put my first number of the bold number sequence. Thus to find the nth term, for 3 by 3 boxes you do 10 x 12.
My formula did work as it showed the right answers. However I believed it wasn’t consistent enough. So I went back to my old theory about the 10n2 would equal the difference of the box, where n represent the square size, even though it did not work straight away. I believed that I should continue to use the formula just to see if it shows a pattern of any kind
I realised that the sequence did show a pattern when 2 was put in n’s place, the result was 40 instead of 10, when 3 was entered in n’s place in the formula, the result was 90 instead of 40.
I can conclude from this that the formula works with a little error, because instead of showing the sequence that I want, it always showed the next term in the sequence. After many trials I found a solution. Since the answer was always one term away. I checked what would happen if I subtract 1 from the number, that is substituting the n’s place.
1 ← 2 ← 3 ← 4 ← 5 (sequence goes down by 1)
2x2 3x3 4x4 5x5 6x6
10 40 90 160 250
30 50 70 90
20 20 20
Since my sequence increase by 1 each time in the first sequence, therefore it can also decrease by 1 each time, so I came with the idea of this new formula 10(n-1)2, where n represent the box size, and 10 is the grid size e.g. for a 2x2 box, 2 is put in n’s place in the formula.
Reason n is squared is because the n = n. e.g. there were 2 rows and 2 columns for 2by2 box, there were 3 rows and 3 columns for a 3by3 box.
Hence the formula makes perfect sense, because 3 is an uneven number and could never give an even when squared so, by subtracting 1 from 3 give 2, which an even number so therefore can give an even number when squared. The same for 4, four is an even number and can never result in a uneven number when it’s squared, so by subtracting 1 from 4, will give an uneven number, and will result in an uneven number when squared.
My overall conclusion is that the formula for any size square box selected any where in a 10 by 10 grid size will be 10(n-1)2
Where n represent the square size. To prove this I will check the product difference for 8 by 8 boxes using my formula
10(8-1)2 = 490
Looking at my previous arithmetic work, I can confidently say that the answer is correct.
To investigate further I have chosen to examine rectangles in a 10 by 10 grid size. I will use the same method as the square box investigation to find the difference of the different size rectangles.
I selected to look at rectangles with constant width of 2, and a length that increased by 1 each time e.g. 2 by 3, 2 by 3, 2 by 4, and 2 by 5.
Aim: find the product difference of different size rectangle boxes
First I will draw a box rectangle box, which has 3 rows and 2 columns. I’ll then find the product of the numbers cross opposite each other in the rectangle.
The product is found by multiplying the top left number with the bottom right number in the box. Now do the same thing with the top right and the bottom left as shown in the diagram.
1 2 3
11 12 13
21 22 23
Product 1: 1 x 22 = 22
Product 2: 2 x 21 = 42
Now subtract the two products, to find the difference
42 – 22 = 20
I will use this method to calculate the difference of more 2 by 3 boxes, chosen randomly in the big 10 by 10 grid.
Now I will prove that the difference is 20 algebraically, by selecting any 2 by 3 rectangle box in the 10 by 10 grid. Replace one of the numbers in the 2x3 box with N
→
Difference is found by subtracting the two products (bold numbers)
N (N + 21) – (N + 1) (N + 20)
N2 +21N – N2 + N +20N + 20
N2 +21N – N2 + 21N + 20
Difference = 20
Conclusion: the difference for a 2x3 rectangle box is always 20 regardless, whether the rectangles are horizontal or vertical.
N (N + 31) – (N + 1) (N + 30)
N2 +31N – N2 + 1N +30N + 30
N2 +31N – N2 + 31N + 30
Difference = 30
Conclusion: the difference for a 2x4rectangle box is always 30 regardless, whether the rectangles are horizontal or vertical.
N (N + 41) – (N + 1) (N + 40)
N2 +41N – N2 + N +40N + 40
N2 +41N – N2 + 41N + 40
Difference = 40
Conclusion: the difference for a 2x5 rectangle box is always 40 regardless, whether the rectangles are horizontal or vertical.
Once more my results formed a pattern. Now I can predict the product difference between the different boxes.
2x3 2x4 2x5 2x6 2x7 2x8 2x9 2x10
20 30 40 50 60 70 80 90
10 10 10 10 10 10 10
Using the method above, I calculated the box difference for the following rectangles. 3 by 2, 3 by 4, 3 by 5, and then 4 by 2, 4 by 3, 4 by 5, and then 5 by2, 5 by 3, 5 by 4. I will ignore 3 by 3, 4 by 4, and 5 by 5. Since these are not rectangles.
3 by 2 = 20 4 by 2 = 30 5 by 2 = 40
3 by 4 = 40 4 by 3 = 60 5 by 3 = 80
3 by 5 = 60 4 by 5 = 90 5 by 4 = 120
The rectangles with a constant width of 3 increases by 20 each time, so to predict the next I would just and 20 each time to the previous term. E.g. 3 by 5 = 60 to find the box difference for a 3 by 6 you add 20 to 60, i.e. 20 + 60 = 80.
I used this principle of adding the constant difference to the previous term in the sequence to find the next term in the sequence. I decided that tabulation was the best way presents my data.
Now I will find a constant formula for the rectangles boxes as I did for the squared boxes. Formula to find the nth for a quadratic sequence is an2 + bn + c. I decided to use my formula 10 (n-1)2
Even though, my sequence wasn’t a quadratic sequence. Looking carefully at the formula I noticed that it was squared. I didn’t need to square my numbers because they were not the same numbers in the columns and rows. Therefore by rewriting the formula I eliminated the square.
10 (n-1)2 = 10 (n-1) (n-1)
I tested if my new theory worked, by substituting the value of constant width of 2 (row) and a length of 3 (column) into my formula. 10 (2-1) (3-1) = 20. This the correct answer, because when I did the arithmetic work for 2x3 boxes I also got 20 as the product difference. The formula however needs to be modified, since n has two values, one constant number, and a changing number.
The formula for rectangle boxes is 10 (m-1) (n-1), where m represent the number of columns and n represent the number of rows. E.g. 2x3, the 2 would be the m and the 3 would be the n in the formula. To find the product difference for any size box in any size grid. You use this formula g (m-1) (n-1), where g represents the grid size and m represents the number of columns and n represent the number of rows.
