Borders - a 2 Dimensional Investigation.

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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

4x4 shape = 25 squares

5x5 shape = 41 squares

6x6 shape = 61 squares

I now have enough data to analyse my results:

        To work out which formula to use I will now put my results in a table showing the differences between the numbers:

n        tn        1st difference         2nd difference

                

1        1 

                        4

2        5                                        4

                        8

3        13                                        4

                        12

4        25                                        4        

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                        16

5        41                                        4

                        20

6        61

I can clearly see that the 2nd difference is the same, which means that my sequence is a quadratic sequence. The formula to find a quadratic sequence is tn = a+ bn + c, where tn equals the total number of squares.

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

When n = 2:                4a + 2b + c        = 5        (2)

When n = 3                9a + 3b + c        = 13        (3)

(2) - (1)                3a + b                = 4        (4)

(3) - (2)                5a + b                = 8        (5)

(5) - (4)                2a                = 4

(÷ 2)                        a                = 2

(substitute into (4))        6 + ...

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