# Investigation into the progression of patterns in 3d shapes.

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Introduction

Borders Coursework

I am continuing my investigation into the progression of patterns further, in that, I am no longer working with 2D shapes, but instead 3D ones. I have drawn the first four patterns on separate isometric paper and the pattern number relates to its nth term, i.e. pattern 2 is the same as n = 2.

In order to draw the shapes I have drawn them in separate layers, to make it easier to count the total number of squares in each pattern.

When drawing each pattern, I added 1 square to each free side of the previous pattern. For instance, pattern 1 is a cube which has 6 sides; therefore I added 6 cubes to it.

On the following pages you will see the diagrams that show how the 3D patterns are built up. As shown by my drawings, each pattern’s number of layers increases. If we look at each individual layer we can see a significant link between the 2D and 3D patterns.

First of all I looked at the centre layer, which is built up in exactly the same way as the 2D patterns.

Middle

My table of results shows that two identical layers are added each time and also, as you look down each column you can see the progression of each individual layer. As I mentioned earlier, it progresses the same as the 2D patterns, which is how I worked out the number of cubes in the each layer of pattern 5. This is what the bottom line of my pyramid table looks like:

D | C | B | A | Centre Layer | A | B | C | D | Nth term |

1 | 5 | 13 | 25 | 41 | 25 | 13 | 5 | 1 | N = 5 |

I can also work out the number of cubes in the centre layer algebraically by using the formula:

Nth term = 2n – 2n + 1

(that being the formula for the 2D pattern)

In order to work out the number of cubes in the centre layer I must use this formula with n being = 5:

2n – 2n + 1

= (2x5x5) – (2x5) + 1

= 50 – 10 + 1

= 41 cubes.

Total Number of Cubes in Each Pattern

Pattern no. (n=) | Total no. of cubes |

1 | 1 |

2 | 7 |

3 | 25 |

4 | 63 |

5 | ? |

Now that I know how many cubes each layer contains I could either draw pattern 5, or simply add the bottom line of my pyramid table.

After drawing this shape and assuming that it is in 3D I can see that it has a total of 129 cubes,

1 + 5 + 13 + 25 + 41 + 25 + 13 + 5 + 1 = 129 cubes.

Conclusion

37a + 7b + c = 38

24a + 2b = 28 (this can be simplified to 12a+b=14)..G)

The equations that are green are the ones I will use for the fourth step in finding the nth term. I can now work out the 3rd group of differences:

Equation F) – E) 9a + b = 10

6a + b = 6

3a = 4 (this can be simplified to a = 4/3)

Equation G) – F)12a + b = 14

9a + b = 10

3a = 4 (this can be simplified to a = 4/3)

I now have a number for ‘a’ in the cubic polynomial and now if I use equation 1) I can find ‘b’.

(6 x 4/3) + b = 6

8 + b = 6

b = -2

Now if I put ‘a’ and ‘b’ into equation A I will be able to find the value of C.

(7 x 4/3) + (3 x -2) + c = 6

28/3 – 6 + c = 6

28/3 – 18/3 + c = 18/3

c=8/3

By putting ‘a’, ‘b’ and ‘c’ into equation 1. I can work out the value of ‘d’

4/3 – 2 + 8/3 + d = 1

4/3 – 6/3 + 8/3 + d = 1

6/3 + d = 1

2 + d = 1

d = -1

Now that I have the values for ‘a’, ‘b’, ‘c’ and ‘d’ I can substitute them into the polynomial:

Nth term = 4/3n - 2n + 8/3n – 1

Now I need to check the formula to make sure it is correct. I will use n=2.

N=2

Second term = (4/3 x 2 ) – (2 x 2 ) + (8/3 x 2) – 1

= 32/3 – 8 + 16/3 – 1

= 48/3 – 9

= 16 – 9

= 7

As the total number of cubes in pattern 2 is 7 this formula appears to be correct.

This student written piece of work is one of many that can be found in our GCSE Hidden Faces and Cubes section.

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