# A* grade maths GCSE borders coursework investigation

## Introduction

**Borders Coursework**

The diagrams below show a square that has been surrounded by more squares to make a cross shape. The cross shape has then been surrounded by more squares to make an even bigger cross shape.

The first diagram consists of 1 square, the second diagram consists of 5 squares while the third diagram has 13 squares in total.

The next cross shape is always made by surrounding the previous cross shape with a certain number of squares.

I am going to investigate, using algebra, how many squares would be needed to make any cross shape built up in this way.

Once I have found the formula for the first part of my investigation, I will extend my investigation to three dimensions and find out the formula for the number of squares added each time.

When I find out what the formulas are in each case, I will check them and make sure they work.

## Part 1

I am going to draw some diagrams to help me find out how many squares are in each diagram. I will then make a table so I can see clearly see the number of squares in each diagram.

I have surrounded each diagram with a different colour of squares because it makes it clearer which layer is the new one each time.

The sequence is a quadratic sequence because the second difference is constant. I am going to find out the formula using a quadratic equation.

In general, a quadratic equation has the form: __y = ax____2____ + bx + c__ where x is the number.

I am now going to work out what the formula is using the __quadratic equation.__

1) 1 = a + b + c - this is 12 = 1, x a which equals a, + 1 x b which equals b + c

2) 5= 4a + 2b + c - this is 22 = 4, x a which equals 4a, + 2 x b which equals 2b + c

3) 13 = 9a + 3b + c - this is 32 = 9, x a which equals 9a, + 3 x b which equals 3b + c

I am going to solve these equations by means of using simultaneous equations. I only need to use 3 equations because there are only 3 unknowns: a, b and c.

If 2a = 4, a = 2

So a = 2, using 5a + b = 8 - 5 x 2 = 10, + b = 8, this means b = __-2__

__ __

Then using a + b + c = 1

2 – 2 + c = 1 so __c = 1__

__ __

__2 – 2 + 1 __

__ __

In general if the quadratic equation has the form __y = ax____2____ + bx + c__

Then the formula to find out the numbers in the sequence is __y=2x____2____ – 2x + 1__

The formula for the 2 dimensions is __2x____2____ – 2x + 1__

This can also be written as **x****2****(x-1)****2**

To check the formula I will work out the 1^{st} and 2^{nd} terms in the sequence.

Both these result are right so my formula is correct.

I am also going to find out what the 7^{th} and 8^{th} terms of the sequence are and see if they fit in with the number pattern of the sequence.

## Three dimensions

**Three dimensions**

I am now going to extend my investigation into three dimensions.

So diagram 1 will consist of 1 cube, diagram 2 will consist of 2 x diagram 1 ( above and below) and 5 cubes in the middle while diagram 3 will consist of 2 x diagram 2, 2 x diagram 1 and 13 cubes.

The diagrams below show three dimensions,

There is a common difference, the first and second differences are not the same but the 3^{rd}

difference is. I think the formula for the sequence contains x3 because I am making it 3D.

As before the number in front of x3 is half the constant difference, so half of 8 is 4.

Therefore the formula contains 4x3

Because the equation to find out the 2D sequence was y = ax2 + bx + c

I think the equation to find out the 3D sequence will be cubic because it is 3D.

The general equation to find out the formula is __y= ax____3____ + bx____2____ + cx + d__

Unlike in the first part of my investigation where I only had to use 3 equations I now have to use 4 equations to work out the formula because there are 4 unknowns: a, b, c and d.

Using the table –

1) a + b + c + d = 1 - this is 13=1, x a which = a, + 12 = 1, x b = b + 1 x c = c + d

2) 8a + 4b + 2c + d = 7 - this is 23=8, x a which = 8a, + 22 = 4, x b = 4b + 2 x c = 2c + d

3) 27a + 9b + 3c + d = 25 – this is 33=27, x a which = 27a, + 32 = 9, x b = 9b + 3 x c =3c + d

4) 64 a + 16b + 4c + d = 63 – this is 43=64, x a which = 64a, + 42 = 16, x b = 16b = 4 x c = 4c + d

Like the equation in the first part of my investigation, I am going to solve these by using simultaneous equations.

I have now found both, the formula for 2 dimensions and the formula for 3 dimensions.

Two dimensions was relatively easy as the second difference was constant. In the 3D sequence it wasn’t until the third difference until it was constant. This gave me the idea, as in the first part of my investigation I worked out that the formula contained x2 because the 2^{nd} difference was constant, I thought that the formula for 3D will have x3 in it because the 3^{rd} difference is constant. The number in front of x2 in the first part of my investigation was half the constant difference which was 4 so I thought the number in front of x3 will also be half the constant difference which was 4.

I also noticed, that the result of simultaneous equations a, b and c matched the pattern of the second difference of the number pattern.

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