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  • Level: GCSE
  • Subject: Maths
  • Word count: 2608

Investigating the number of patterns in a certain grid.

Extracts from this document...

Introduction

COURSEWORK (PAGE 1)

Introduction

This piece of coursework is about investigating the number of patterns in a certain grid. The grid I am starting off with is 10x10. The rule or pattern I have to investigate is whether when you multiply the opposite corners in a 2x2 box, you can find a pattern or link between the two answers.

I will also investigate to see if there are any patterns when I multiply the opposite corners of a box but using a larger grid (increase the row length) or increase the size of the square (3x3etc.).

I will always use algebra to check if my assumptions are correct.

I will also see what happens if I do not use a square but use other shapes such as a rectangle or a T-shape. I will see whether the rule still applies and I will generalise the rule by using algebra.

Finally I will invent my own rule and if it is successful I will generalise using algebra

Here is a list, in order, of the plans of my investigation:

1 check if the rules are universal

Do the arithmetic

2 generalise using y         y+1

   Do the algebra    y +7     y+8

3 change the size of the grid (e.g. row 1-10)

   Do arithmetic

   Do algebra (row 1-n)

4 change the size of the square y         y+1     y+2

   Do the arithmetic                          y+7     y+8     y+9

   Do the algebra                              y+14   y+15   y+16

5 generalise the square y………………y+r

   Do the algebra (nxn)        y+5

   Square

6 extensions

...read more.

Middle

This table shows the difference between the two results you get when you multiply the opposite corners, but it also shows the difference on different size squares. As you can see from the table there is a distinct pattern.

By recognising that the differences are the square numbers multiplied by 10 I was able to deduce what a 5x5 and 6x6 square would give.

If you want to work out a formula for any square (nxn) then you have to realise that it is the side length (n) minus 1 then you square it and finally multiply by 10. You are then left with the formula (n-n)^2 x10.

Now I will change the length and height of the grid to generalise the formula more. The grid is now 13 by 8.

1     2     3     4     5     6     7     8     9     10     11     12     13

14  

27

40

53

66

79

92

These are the examples I am going to use:

14   15                    44   45                  37   38

27   28                    57   58                  50   51

14x28 =392           44x58 =2552        37x51 =1887

15x27 =405            57x45 =2565       50x38 =1900

The difference is 13 every time.  

I am now going to use algebra to check my answer.

Y         y+1                 This is the generalised version of the 2x2 grids.

Y+13   y+14

Y(y+14)         =y +14y

(y+13)(y+1)   =y +14y+13

The algebra proves that the difference is 13.

I will now investigate 3x3 squares.

14    15    16                 44    45    46                37    38    39

27    28    29                 57    58    59                50    51    52

40    41    42                 70    71    72                63    64    65

14x42 =588                  44x72 =3168               37x65 =2405

16x40 =640                  70x46 =3220               39x63 =2457

Difference in every case =52

I am going to use algebra to check my answer.

Y           y+1        y+2                             y(y+28) =y +28y

Y+13     y+14      y+15                           (y+26)(y+2) =y +28y+52

Y+26     y+27      y+28                                                    

After the equations cancel each other out the only difference left is +52.

...read more.

Conclusion

2x2                                                          4R

3x3                                                          16R

4x4                                                          36R

NxN                                                         R((N-1)^2 x4)

I will now attempt to find a formula for any rectangle where the numbers go up by two.

N

              Y                               -                         y+2(N-1)

M           -                                -                             -

              Y+2(M-1)R              -                          y+2(M-1)R+2(N-1)

Y x y+(2(M-1)R+3(N-1))         =y^2 +2MRY-2RY+2YN-2Y

(y+2MR-2R) x (y+2N-2)          =y^2 +2MRY+2NY-2RY-2Y+4RMN-4RM-4RN+4R

Difference =4RMN-4RN-4RM+4R =R(MN-N-M+1)x4 =4(R(M-1)(N-1))

The number two represents how much the numbers go up each time. It van be replaced by the letter H IN the formula to give the complete formula to the problem as:                    H^2(R(M-1)(N-1))

K =number the numbers go up by each time

R =row length of the total grid

M+N =dimensions of the rectangle inside the grid

I will test this formula:

K =3                 R=5               M=2             N=3

3                             6

18                           21

33                           36

Because K was 3 that meant the numbers went up by three every time.

3x36 =108

6x33 =198                             difference =90

K^2(R(M-1)(N-1)) =9(5(1)(2)) = 9x10, =90

The formula appears to work.

I have now completed the investigation trying as many different things as possible and ending up with a final formula using numerical examples and algebra throughout to check what I have done. Of course after I have generalised for any square or rectangle it is only relevant to my peticular grid. That is why at one stage I used 3 examples instead of two as my third example went off the grid as I testes 4x4 squares.

I believe I have carried out a successful investigation.                                                                                        

...read more.

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