# My job is to investigate how many squares would be needed to make any cross shape like this build up in the same way.

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Introduction

Adeel Younis, Math’s Coursework Borders, Mr. Ekwalla, 10A, 10G 5/10/2007

Borders: Part 1

My job is to investigate how many squares would be needed to make any cross shape like this build up in the same way.

Below are diagrams of the cross shape pattern to the 8th sequence:

Here is my table of results, with the number of black, red and total squares

Black Squares | 4 | 8 | 12 | 16 | 20 | 24 | 28 | 32 |

Red Squares | 1 | 5 | 13 | 25 | 41 | 61 | 85 | 113 |

Total Squares | 5 | 13 | 25 | 41 | 61 | 85 | 113 | 145 |

Now so I can find an equation which will tell me how many squares there will be in each sequence, I will find the differences for the black squares first.

To find the equation I will need bits of information, the equation Is an+b, where a is the difference and b is the 0th term, e.g. the first term here is 4, to find the 0th

Middle

2nd Step find the equation, a=2 c=1 now to find b

1)Un=an²+bn+c

2)5= an²+bn+c

3)5=2×1²+b×1+1

4)5=2+b+1

5)5=3+b

6)5-3=3+b-3

7)2=b

Let’s test it; term 1 is 5, 2×1²+2×1+1=5. I’m right again! Now we have the equations for the black, white and total squares

Black Squares= 4n+0

Red Squares = n=2n²+-2n+1

Total Squares = n=2n²+2n+1

Borders: Part 2

Now to expand my investigation into the 3rd dimension. Please find the sheet with my drawing of the 3D shapes on, after you have seen that continue reading

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Here is where maths becomes complicated, the drawings on the last page are the cross shaped in the 3rd dimension, and here is the table showing how many cubes each shape has.

Shape | 1 | 2 | 3 | 4 | 5 | 6 |

No. of cubes | 7 | 25 | 63 | 129 | 231 | 377 |

Now I will find the differences

Conclusion

36

85

116

288

S-4/3n³

5

14

27

43

64

89

Here we go now were left with the quadratic equation, now I will find its differences.

Now we can find b, c and d. b, c and d are basically the a, b and c in the quadratic equation, in the same way we have to find c last and b and c first, b=½ of the second difference, in this case 2, d= 0th term, in this case 1, now we substitute b and d into the equation to find c:

1)Un=bn²+cn+d

2)27= an²+cn+c

3)27=2×3²+3c×3+1

4)27=18+3c+1

5)27=19+3c

6)27-19=19+3b-19

7)8=3b

8)

9)2=c

Now we have c, the full equation is n³+2n²+n+1. Our journey doesn’t end here, we still need to test it, let’s go ahead. We will test it with the 1st term, the 1st term is 7, ×1³+2×1²+×1+1=7 Hooray, the equation works.

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