# Borders - a 2 Dimensional Investigation.

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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 = 2n² - 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

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 = an³ + bn² + 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 = 1n³ -2n² + 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.

Page

This student written piece of work is one of many that can be found in our GCSE Consecutive Numbers section.

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