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.