Un = 2n2 – 2n + 1

This is correct.

Growing Hexagons

I will now repeat my investigation, and change the original shape of the square to hexagons, and try to find the formula as before.

I shall start by finding the width of each hexagon.

I have found the 1st difference to be constant; therefore the formula will be linear.

I have found the formula to be W=2n-1

I will now try to find the number of hexagons with in a pattern

Table of results:

The 2nd difference is constant; therefore the equations will be quadratic. The general formula for a quadratic equation is an2 + bn +c. The coefficient of n2 is half that of the second difference

Therefore so far my formula is: 3n2 + [extra bit]

I will now attempt to find the extra bit for this formula.

From my table of results I have found the formula to be 3n2 + 3n + 1

I will now check my formula by substituting a value from the table in to my formula:

E.g. n=3

Un=3(3) 2 -3(3) + 1 = 19.

For my table of results I can see that my formula is correct after verification.

I have chosen to justify my formula using 3 simultaneous equations rather that geometrically like the previous investigation.

The standard quadratic formula » U = an2 + bn + c

I will now substitute values for U and n

7 = a (2) 2 + 2b + c (1)

19 = a (3) 2 + 3b + c (2)

37 = a (4) 2 + 4b + c (3)

If we then subtract equation (2) from (1) we get

12 = 5a +b (4)

When we do the same between (3) and (2)

18 = 7a + b (5)

(5) – (4) = 6 = 2a

Therefore a = 3

To work out b we must insert the value of a in to equation (4)

(4) » 12 = 5 (3) + b

12 = 15 + b

12 – 15 = b

Therefore b = - 3

To work out c we must insert the values of a and b in to (1)

(1) » 7 = 4 (3) + 2 (-3) + c

7 = 12 –b +c

7 – 12 + 6 = c

1 = c

Therefore c = 1

U = an2 + bn + c

» U = 3n2 + 3n + 1

This matches my original formula

## 3D Shapes

I will now evolve my investigation by looking at a 3D cube. Like the growing squares I will start with 1 and add an additional shape to each available side. To work out and draw each shape I will decompose it in to layers.

Table of results:

The 3rd difference is constant; therefore the equations will be cubic. The coefficient of n3 is 1/6 of the 3rd difference.

I will now attempt to find the extra bit for this formula.

The 2nd difference is constant; therefore the equations will be quadratic. The general formula for a quadratic equation is an2 + bn +c. The coefficient of n2 is half that of the second difference

Therefore so far my formula is: 4/3n2 - 2n² [extra bit]

Therefore my formula is:

4/3n³ – 2n² + 8/6n – 1

I will test this formula out on a pattern previously drawn out and recorded.

Un = 4/3n³ – 2n² + 8/6n – 1

36 – 18 + 4 – 1

= 25

This is correct

I will now and justify the formula by using simultaneous equations.

Let C= An³ - bn² +cn – d

When n = 1, c = 1. Therefore:

1= a +b + c + d (1)

When n = 2, c = 7. Therefore:

7 = 8a + 4b + 2c + d (2)

When n = 3, c = 25. Therefore:

25 = 27a +9b + 3c + d (3)

When n = 4, c = 65. Therefore:

63 = 64a + 16b + 4c + d (4)

If we then subtract equation (2) from equation (1) we get

6 = 7a + 3b + c (5)

If we do the same again to (3) and (2), we get

18 = 19a + 5b + c (6)

Then to (4) and (3)

38 = 37a + 7b + c (7)

Then to (6) and (5)

12 = 12a + 2b (8)

Then again to (7) – (6)

20 = 18a + 2b (9)

And finally to (9) and (8)

A = 8/6 = 4/3

Substitute into 8

12 = 16 + 2b

-2 = b

Substitute into 5

6 = 9 1/3 – 6 + c

2 2/3 = c = 8/3

Substitute into 1

1 = 4/3 + (-2) + 8/3 + d

-1 = d

Therefore my formula is as follows;

C=4/3n³ – 2n² + 8/6n – 1

This is correct.

John McLaughlin –

Feel free to e-mail me on ways to improve this essay it would be a great help.