# Maths -Borders

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Introduction

Now I will draw two different cross shapes from its basic one and see whether any pattern can be observed between the given cross shape and the following:

I will choose the following basic diagrams.

Diagram 2

Diagram 1

Pattern 1

Dark Squares=3

Blank Squares=7

Total Squares=10

Pattern 2

Dark Squares =10

Blank Squares=11

Total Squares=21

Pattern 1

Dark Squares = 5

Blank squares = 10

Total squares = 15

Pattern 2

Dark Squares=15

Blank Squares=14

Total Squares=29

Observations:

- Pattern 1 in diagram 1 consists of an even number of squares while the dark squares in the same pattern are odd and the blank squares are also odd, I noticed that the sum of two odd figures is an even value.
- In pattern 2 in diagram 1 an odd number of squares was added to an even number hence the total value is odd.
- In pattern 1, diagram 2 the total number of squares, the number of blank squares and the number of dark squares are all multiples of 5.
- In pattern 2, diagram 2 the difference between dark squares and blank squares is odd and the difference between the blank squares and total squares is also odd.

Middle

6 61 20

The numbers of squares added are the multiples of four and occur in this order:

4 * 0 = 0

4 * 1 = 4

4 * 2 = 8

4 * 3 = 12

4 * 4 = 16

4 * 5 = 20

Therefore I can see a sequence in the order of these differences and the common difference between them is four.

Therefore the difference between the number of squares in the 6th and the 7th level will be 24 and the difference between the 7th and 8th will be 28.

Therefore 61 + 24 = 85

The 7th level contains 85 squares

85 + 28 = 113

Hence, the 8th level will contains 113 squares

7 | 85 | 24 |

8 | 113 | 28 |

Level 7

Level 8

The 8th level comprises of 113 squares, this means that there are 28 more squares.

Here I will prove my forecasts by extracting the same conclusion in diagrammatical way;

Level 4

Level 5

The line in the fourth level crosses 3 squares on the outer layer of the figure. The lines all cross through the blank squares only. In the diagram above, the four edge squares are crossed twice by the lines.

Conclusion

n = the number of the level of the diagram

b and c need to be found in the equation to solve the equation.

Constant difference = 4

Therefore ‘a’ = ½ x 4

a = 2

When n = 1

If n=1, 1= 2(1)2 b+(1)+c

1= 2+b+c

b+c = 1-2= -1

if n=2, 5=2(2)2 + b(2)+c

5=8+2b+c

2b+c= 5-8= -3

Solve simultaneously= equation 2 -1

= 2b+c= -3 1

-b+c= -1 2

b=-2

2b+c= -3

2(-2) +c= -3

-4 + c= -3

C= -3+4= 1

a = 2

b= -2

c = 1

Consequently, Un = an2 +bn +c

UN= 2n2 -2n +1

Verify in level 4

Un = 2n2 + -2n + 1

2(4)2 + -2(4) + 1 = 25

2 x 16 – 8 + 1 = 25

32 - 8 + 1 = 25

32–7= 25

25 = 25

The previous calculation proves that the formula is accurate. Now I will prove it diagrammatically.

1

3

5

1 3

The first row has 1 square and therefore I have marked the row by an arrow.

Level | Total number of squares | Break up |

1 | 1 | 1 |

2 | 5 | 1+3+1 |

3 | 13 | 1+3+5+3+1 |

4 | 25 | 1+3+5+7+5+3+1 |

5 | 41 | 1+3+5+7+9+7+5+3+1 |

6 | 61 | 1+3+5+7+9+11+9+7+5+3+1 |

Therefore I can get the formulae 4(n – 1) = the added squares.

Here I will further my endeavors to the anlysation and investigation to Three Dimensional shapes.

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