Part 1: 2D Shapes:

1st term: 2nd term:

3rd term: 4th term:

5th term: 6th term:

7th term: 8th term:

9th term:

Part 1: 2D Shapes:

Table to show the amount of squares in total and the amount of extra squares used for the 2D drawing:

Working out to show the equation (nth term) for the Extra Squares used.

0 4 8 12 16 20 24 28 32

+4 +4 +4 +4 +4 +4 +4 +4

Therefore the first part of the equation is 4n.

To find out the next part of the equation, I will need to see how the number 4 is related to the value of ‘n’ and work out the difference, to find out the remaining equation.

When n= 1 the number of extra squares is 0.

So: 1*4=4

4+/- x= 0

x= -4

General equation for Linear sequences: an+ b

So the equation is: 4n-4.

To make sure the equation is correct; I will check it using values of ‘n’.

When n = 2, the number of extra squares is 4.

So: 2 * 4 = 8

8 – 4 = 4

The answer matches the answer in the sequence so the equation is correct, but I will check again to be sure.

When n = 5, the number of extra squares is 16.

So: 5 * 4 = 20

20 – 4 = 16

The equation is correct as both times the answers have matched those in the sequence.

Working out to show the equation (nth term) for the Total Squares used.

1 5 13 25 41 61 85 113 145

+4 +8 +12 +16 +20 +24 +28 +32

+4 +4 +4 +4 +4 +4 +4

- 2a = 4 2) 3a + b = 4 3) a + b + c = 5

a = 2 6 + b = 4 2 + 2 + c = 5

b = -2 c = 1

General equation for Quadratic sequences: an2 + bn + c

So the equation for this sequence is: 2n2 - 2n + 1

To check that the equation is correct I will use values of ‘n’ to make sure.

When n= 4, the total number of squares used are 25.

So: 2 * (42) – (2*4) + 1 = 25

This shows the equation is correct. I will however, check again to be sure of this.

When n = 9, the total squares used are 145.

So: 2* (92) – (2*9) + 1 = 145

This tells that the equation works and is therefore correct.

Part 2: 3D Shapes:

(The 3D drawings are on the triangle dotted paper attached.)

Table to show the amount of squares in total and the amount of extra cubes used for the 3D drawings:

Working out to show the equation (nth term) for the Total Cubes used.

Sequence:

1 7 25 63 129

+6 +18 +38 +66

+12 +20 +28

+8 +8

Cubic sequence :

1 8 27 64 125

+7 +19 +37 +61

+12 +18 +24

+6 +6

General equation for Cubic sequences: an3 + bn2 + cn + d

To find ‘a’ in the formula:

a= 3rd difference in sequence

3rd difference in cubic sequence

So: 8/6 = 4/3 =1⅓

Therefore the first part of the equation is: 1⅓ n3 + bn2 + cn + d

New sequence:

-1/3 -11/3 -33/3 -67/3 -113/3

-10/3 -22/3 -34/3 -46/3

-12/3 -12/3 -12/3

General Quadratic equation: bn2 + cn + d

2b= -12/3 (-4)

b= -6/3 (-2)

3b + c= -10/3

-18/3 (-6) + c= -10/3

c= 8/3 (2 2/3)

b + c +d= -1/3

-6/3 + 8/3 + d = -1/3

d= -1

Remaining equation for total number of cubes used in any 3D cross- shape: -2n2 + 22/3n – 1

Full equation to work out the total cubes needed to make any 3D cross- shape:

11/3n3 -2n2 + 22/3n – 1

To check the formula works, I will test it using values of ‘n’.

When n= 1, the total cubes used is 1.

So: (1⅓ * 13) – (2 * 12) + (22/3 * 1) – 1

= 1⅓ - 2 + 22/3 -1 = 1

This shows that the formula works, but I will check again, in case I have made a mistake.

When n= 5, the total cubes used are 129.

So: (1⅓ * 53) – (2 * 52) + (22/3 * 5) – 1

= 125⅓ -50 +72/3 -1 = 129

This proves that my formula is correct because the answers match those in my sequence.

1st term 2nd term 3rd term

4th term 5th term

Justification of 2D shapes: