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• Level: GCSE
• Subject: Maths
• Word count: 1104

# Borders - a 2 Dimensional Investigation.

Extracts from this document...

Introduction

Mel Spencer 10L

Maths Coursework

BORDERS

The shape on the left is a dark cross shape that has been surrounded by white squares to create a bigger cross shape. The shape consists of 25 small squares in total. To make the next shape in the sequence, white squares are added to the existing shape, creating a bigger cross shape made up in the same way.  For my investigation, I am going to work out how many squares would be needed to make up any cross shape built up in this way by working out a general formula. Then I am going to extend my investigation to 3 dimensions.

I intend to use structure to work out a general formula.

2 Dimensional Investigation

I am going to start by building a 1x1 cross-shape and adding on the borders. The complete shape will be known as the cross-shape (this includes the border squares).

I have decided to define this shape as a 1x1 shape.

1x1 shape = 1 square

2x2 shape = 5 squares

3x3 shape = 13 squares

Middle

(- 6)                        b                = -2

(substitute into (3))         18 - 6 + c        = 13

12 + c         =13

(- 12)c                = 1

Therefore my formula is: tn = 2- 2n + 1

Independent Check:

When n = 4, tn should equal 25

2(4²) - 2(4) +1

32 - 8 + 1        =25

My formula is correct for this sequence, but to make sure my formula is accurate, I am now going to prove it using structure.

When I look at any of the cross shapes, I see that they can be split up into four triangles, with one square left in the middle:

The formula for the sequence of triangular numbers (1, 3, 6, 10 etc.) is n(n+1)

2

n(n+1)  = 1(1+1)  = 1

2              2

n(n+1)  = 2(2+1)  = 3

2        2

n(n+1)  = 3(3+1)  = 6

2        2

As there are 3 squares in each of the triangles, and 3 is the 2nd number in the triangular numbers sequence, I know that if I use the standard formula for finding the triangular numbers, it will not work, because the cross shape is 3rd in the sequence, not 2nd. This means that I have to use a different formula.

If instead of using n(n+1) I use

Conclusion

60;1

6

2        7                                        12

18                                                8

3        25                                        20

38                                                8

4        63                                        28

66                                                8

5        129                                        36

102

6        231

The 3rd difference is the same, so therefore tn = a+ b+ cn + d.

When n = 1        a + b + c + d                 = 1        (1)

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

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

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

(2)-(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

(÷6)        a                        =8/6 = 1

(substitute into (9))        24 + 2b                        = 20

(-24)        2b                        = -4

(÷2)        b                        = -2

(substitute into (5))        9 - 6 + c                        = 6

3 + c                        = 6

(-5)        c                        = 2

(substitute into (4))        85 - 32 + 10 + d        = 63

64 + d                        = 63

(-56)        d                        = -1

Therefore my formula is: tn = 1 -2+ 2n + 1

Independent Check:

When n = 5, tn should equal 129

1 (5³) - 2(5²) + 2(5) + 7

166 - 50 + 13 - 1        = 129

My formula is correct for this sequence.

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