4 16 41
4
5 20 61
4
6 24 85
4
7 28 113
4
8 32 145
The difference between the white squares after each pattern is 4; this shows it is a multiple of 4 and has a regular pattern and maybe a linear equation. The first formula I will try to find is the formula for the surrounding white squares.
Trying for a Formula- white squares
I will show he first 4 patterns as an example. Noticing that the white squares have a regular difference for indicates it to be a linear equation. By noticing the regular difference I have worked out that you have to multiply the pattern number by 4.
Pattern Number White Squares
1 x 4 4
2 x 4 8
3 x 4 12
- x 4 16
n x 4 4n
So in result the general formula for this would be n x 4 or 4n. I will prove this formula to be correct by choosing a formula at random. I have chosen the number 10. 4n= 4 x 10= 40 white squares.
I am now going to search for the formula the formula for the dark squares. I will the information from the result table and differences (from pg2) I have found that the number of shaded squares equals to the total from the previous squares. To see this look at page 2. I have decided to draw a new table similar to that of the previous page.
Pattern Total Squares Shaded Squares White Squares Difference in
Squares
1 5 1 4
4
2 13 5 8
8
3 25 13 12
12
4 41 25 16
Throughout the investigation I will use the following symbols for the formulae, these will not change. n = pattern number, d= dark squares, w = white squares.
Trying for a formula for shaded squares
I have found out the general formula to find the number of shaded squares which is 2nsquared – 2n + 1, because this is a quadratic form of ax +bx+c, I found a second difference between the pattern of shaded squares. Because the number total for shaded squares are odd you have to add one to the equation. I have found that you have to multiply the pattern number by 2 and square it to get the white squares and then minus 2 which is again multiplied by the pattern number and add 1 to make the number odd.
Example 1:- 2n – 2n + 1 Example 2:- 2n – 2n + 1
Pattern 5: 2 x 5 – 10+ 1 Pattern 6: 2 x 6 – 12 + 1
= 50 – 10 + 1 = 72 – 12 + 1
= 41 = 61
41+20= 61
I have drawn a diagram with 41 shaded squares
and have surrounded them with white squares.
Which I have shaded in a more lighter colour to find
the next pattern for dark squares. I have also
proved my equation to be correct.
Now I have found the formula for shaded
squares and proved it to be correct I am
going to try to find the formula for the total
number of squares.
Trying for a formula for total number of squares
I will draw the table of like the previous one to help me find the sequence.
Pattern White Squares Shaded Squares Total
1 4 1 5
2 8 5 13
3 12 13 25
4 16 25 41
I have found out that to find the total number of squares you have to add the white squares and shaded squares together.
So in result to I can add the formula for the white squares with the formula to shaded squares to give me the formula for total number of squares.
White Squares = 4n add both together = 2n – 2n +1 +4n
Dark Squares = 2n – 2n +1 simplify it = 2n + 2n + 1
Now I have found out the formula, I am going to predict that the next total number of squares in the pattern will be 61.
For Example:- 2n + 2n + 1 I have proved my prediction and the
Pattern 5 = 2 x 5 +2 x 5 + 1 formula to be correct. It is a quadratic.
= 6
Now I have found all the formulae, I am going to record all the findings in a table from pattern number 1 – 14.
Pattern White Squares Shaded Squares Total
1 x4 4 1 5
2 x4 8 5 13
3 x4 12 13 25
4 x4 16 25 41
5 x4 20 41 61
6 x4 24 61 85
7 x4 28 85 113
8 x4 32 113 145
9 x4 36 145 181
10 x4 40 181 221
11 x4 44 221 265
12 x4 48 265 313
13 x4 52 313 365
14 x4 56 365 421
Conclusion
By carrying out this investigation I have found that by using one single formula I am able to figure out the rest of the remaining formulae and be able to find the patterns without drawing the diagrams.
Drawing simple diagrams at first can make it easier to find the answer for the investigation. Another good way is using different tables which hold a record of your findings take a part in finding an equation by recording the information on the table to justify your results. Using different techniques to tackle a problem is beneficial because of the reason that I drew a number of diagrams and a variety of assorted result tables to explain what I did. I also used different symbols and a plan to explain what I had done.
Equation 2n – 2n + 1 is very similar to equation 2n + 2n +1. I noticed this immediately after I had found the second equation. The difference between the two is the symbol before 2n. One is – (negative) and the other + (positive). This was because I had to add combine two equations together. The formula for white squares 4n and the formula for shaded squares 2n – 2n + 1, which in result gave 2n + 2n + 1.
The first equation I found was a linear equation, as 4n can be written as 4x. Equation 2 is a quadratic as it is in the form ax + bx+ c. the same with equation 3. I am planning to extend this investigation to 3D.
