# Borders Investigation Maths Coursework

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Introduction

Mohammed Sharif 11S Page No: Math’s Coursework

Borders Investigation

Introduction

Below is the starting point of my sequence of cross shapes

Aim: To investigate the sequence of squares in a pattern needed to make any cross shapes built in this way and then extend your investigation to 3 dimensional.

In this investigation I have been asked to find out how many squares would be needed to make up a certain pattern according to its sequence, also to derive algebraic formulae from the sequence each expressing one property in terms of another. I plan to solve my investigation by using a wide variety of mathematical tools such as the nth term and also using formulae which includes the linear and quadratic sequence, formulae will be checked and hopefully proven by different mathematical tools.

My sequence starts with a single white square, and then it’s surrounded by black squares (borders) which will form the next shape. To get the second or third shape the second shape becomes white with a border of black squares around it so in each new cross shape, the previous cross shape can be seen as the area of white squares in the centre. I chose to start from one square because I thought it was creative and will look interesting as it builds up.

Middle

2+1= 3 then I did

3-5 = -2 which is b in the equation

So therefore the equation is complete and will now look like:

tn = 2n² - 2n + 1

Testing the rule

As you can see above I have now got a complete formula which I will check to see if it works

Sequence two

- 2(22) -4 +1
- 2(4) -4 +1
- 8 – 4 +1

= 5 Correct

As you can see the formula is working I am going to try it on some more sequences to be 100% sure that it is fully working and has no errors

Sequence 3

- 2(32) -6 +1
- 2(9) -6 +1
- 18 – 6 +1

= 13 Correct

Sequence 6

- 2(62) -12 +1
- 2(36) -12 +1
- 72 -12 +1

= 61 Correct

The formula that I found seems to be working fine without any errors as proven above. I am now going to use this formula to find the number of squares (tn) in a higher sequence:

Sequence 11

- 2(112) -22 +1
- 2(121) -22 +1
- 242-22 +1

= 221 Correct

To check if I am correct or incorrect I will start from sequence 6 as the 1st difference for every pattern is 4 and the 2nd difference is constant. I can see that +4 is added to each pattern in the 1st difference so to do this I will add the total number of squares with the 1st difference for sequence 6 through its previous number and then I will add the 2nd difference which is contact +4 the same applies to all other sequences, the table below should explain everything

Sequence | Total no of Squares | |

7 | 61+20+4 | 85 |

8 | 85+24+4 | 113 |

9 | 113+28+4 | 145 |

10 | 145+32+4 | 181 |

11 | 181+36+4 | 221 |

Conclusion

When n = 2 a(2)3+b(2)2+c(2)+d =7

8a + 4b + 2c + d =7 (2)

When n = 3 a(3)3+b(3)2+c(3)+d =25

27a + 9b + 3c + d = 25 (3)

When n = 4 a(4)3+b(4)2+c(4)+d =63

64a + 16b + 4c + d = 63 (4)

Simultaneous Equations

For the first I am going to show the full working out, so it’s easier to understand

(2)-(1) 8a + 4b + 2c + d=7

- a + b + c + d=1

= 7a+3b+c =6 (5)

(3)-(2) 19a+5b+c =18 (6)

(6)-(5) 12a+2b =12 (7)

(4)-(3) 37a+7b+c =38 (8)

(8)-(6) 18a+2b =20 (9)

(9)-(7) 6a =8

6a =8

÷6 ÷6

We can’t do it this way as it comes in decimals therefore we have to do it in fractions

8÷2

6÷2 = 4/3 therefore a= 4/3

Substituting a= 4/3 into equation (9) to find b

Which is

18a+2b=20

18x4/3+2b= 20

=72/3+2b= 20

=24+2b=20

24-24+2b=20-24

2b= -4

To leave b on it’s on I have to divide both sides by two to and then that will help me to find b

So that’s

2b/2= b -4/2= -2 therefore b= -2

To find c substitute a = 4/3 and b = -2 into equation (8)

37a+7b+c=38

37x4/3+7(-2)+c=38

=148/3-14+c=38

148 14

3 1

X3

=

148 42 = 106/3

3 3

106/3 + c = 38

106/3 - 106/3 + c = 38 - 106/3

38-(106÷3)= 8/3 so therefore c = 8/3

To find d substitute a = 4/3, b = -2 and c = 8/3 into equation (1)

a+b+c+d=1

4/3-2+8/3+d=1

4-2+d=1

2+d=1

2-2+d=1-2 which is -1 therfore d = -1

Now I have my complete formula= 4/3n3 - 2n2 + 8/3n – 1

Formula 4/3n3 - 2n2 + 8/3n – 1 checking

n 2 = 7

4/3(2)3 – 2(2)2 + 8/3(2) – 1=7

4/3(8) – 2(4) + 16/3 – 1 = 7

32/3 -8 + 16/3 -1 = 7

32/3+16/3= 16

16-8= 8-1= 7 CORRECT

n 5 = 129

4/3(5)3 – 2(5)2 + 8/3(5) – 1=129

4/3(125) – 2(25) + 40/3 – 1 = 129

500/3 -50 + 40/3 -1 = 129

500/3+40/3= 180

180-50= 130-1= 129 CORRECT

My formula is correct, and therefore it should work on all the sequences as proven above.

This student written piece of work is one of many that can be found in our GCSE Hidden Faces and Cubes section.

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