be found out by taking the odd numbers from 1 onwards and adding
them up (according to the sequence). We then take the summation
(å) of these odd numbers and multiply them by two. After doing this
we add on the next consecutive odd number to the doubled total.
I have also noticied something through the drawings I have made
of the patterns. If we look at the symetrical sides of the pattern
and add up the number of squares we achieve a square number.
Attempting to Obtain a Formula Through the Use of the Difference Method
I will now apply Jean Holderness' difference method to try and find a formula.
Pos.in seq. 1 2 3 4 5 6
No.of squar. (c) 1 5 13 25 41 61
1st differ. (a+b) 0 4 8 12 16 20
2nd differ. (2a) 4 4 4 4 4
We can now use the equation
an<sup>2</sup> + bn + c
'n' indicating the position in the sequence.
If a = 2 then c = 1 and a + b = 0 If
2 is equal to b- then b = -2
I will now work out the equation using the information I have obtained
through using the difference method:
1) 2(n -1) (n - 1) + 2n - 1
2) 2(n<sup>2</sup> - 2n + 1) + 2n - 1
3) 2n<sup>2</sup> - 4n + 2 + 2n - 1
4) 2n<sup>2</sup> - 2n + 1
Therefore my final equation is:
2n<sup>2</sup> - 2n + 1
Proving My Equation and Using it to Find the Number
of Squares in Higher Sequences
I will now prove my equation by applying it to a number of sequences
and higher sequences I have not yet explored.
Sequence 3:
1. 2(3<sup>2</sup>) - 6 + 1
2. 2(9) - 6 + 1
3. 18 -5
4. = 13
The formula when applied to sequence 3 appears to be
successful.
Sequence 5:
1. 2(5<sup>2</sup>) - 10 + 1
2. 2(25) - 10 + 1
3. 50 - 10 + 1
4. 50 - 9
5. = 41
Successful
Sequence 6:
1. 2(6<sup>2</sup>) - 12 + 1
2. 2(36) - 12 +1
3. 72 - 12 + 1
4. 72 - 11
5. = 61
Successful
Sequence 8:
1. 2(8<sup>2</sup>) - 16 + 1
2. 2(64) - 16 + 1
3. 128 - 16 + 1
4. 128 - 15 5. = 113
Successful
The formula I found seems to be successful as I have shown on the
previous page. I will now use the formula to find the number of squares in a higher sequence.
So now I wil use the formula 2n<sup>2</sup> - 2n + 1 to try and find
the number of squares contained in sequence 20.
Sequence 20:
2 (20<sup>2</sup>) - 40 + 1
2(400) - 40 + 1
800 - 40 + 1
800 - 49
= 761
Instead of illustrating the pattern I am going to use the method
I used at the start of this piece of coursework. The method in which
Iused to look for any patterns in the sequences. I will use this
to prove the number of squares given by the equation is correct.
As shown below:
2(1+3+5+7+9+11+13+15+17+19+21+23+25+27+29+31+33+35+37) + 39 = 761
I feel this proves the equation fully.
Using the Difference Method to Find an Equation to Establish
the Number of Squares in a 3D Version of the Pattern
Pos.in seq. 0 1 2 3 4 5
No.of squar. -1 1 7 25 63 129
1st differ. 2 6 18 38 66
2nd differ. 4 12 20 28 36
3rd differ. 8 8 8 8
So therefore we get the equation;
anƒ + bn<sup>2</sup> + cn + d
We already know the values of 'n' (position in sequence) in the
equation so now we have to find out the values of a, b, c, and d.
If n = 0 then d = -1 and if n = 1 then d = 1
I can now get rid of d from the equation to make it easier to find
the rest of the values. I will will take n = 2 to do this in the following way:
1st calculation
_ 8a + 4b + 2c + d
a + b + c +d
7a + 3b + c
D will always be added to each side of the equation.
2nd
8a + 4b + 2c = 8 = 4a + 2b + c = 4
2
So then n = 2 8a + 4b + 2c = 8 = 4a + 2b + c = 4
n = 3 27a + 9b + 3c = 26
n = 4 64a + 16b +4c = 64 = 16a + 4b + c = 16
4
To get rid of 'c' I will use this calculation;
_16a + 4b + c = 16
4a + 2b + c = 4
12a + 2b = 12
We can simplify this equation to:
6a + b = 6
My next calculation is below:
N =3 <eth> _27a + 9b + 3c = 26
12a + 6b + 3c = 12 15a + 3b =14
(15a + 3b = 14) ÷ 3
= 5a + b = 4Y
If I use the equation above 6a + b = 6. I can take my latest
equation and subtract it from it to find 'a'.
So, _6a + b = 6
5a + b = 4Y
a = 15
Now that we nave obtained 'a' we can now substitute its value into
the equation to find the other values.
We can now find 'b' by substituting in 15 into the equation as follows:
5(_) + b = 4Y
5(_) = 6Y
b = 4Y - 6Y
b= -2
We have now found values a & b.
I will now attempt to find values c and d by substituting in the two values I now possess.
So we will now sub. in the values to the equation
If we take sequence 2 for our value of n with our present values we will get:
_(8) - 2(4) + c = 8
this can then be simplified to,
_(2) - 2(2) + c = 4
by dividing the equation by two.
We will continue with the calculation using the simplified version of the equation;
To find 'c' we will use the following calculations:
1) _(4) - 2(2) + c = 4
2) 55 - 4 + c = 4
3) 55 + c = 8
4) c = 8 - 55
5) c = 2Y
Therefore my four values are:
a = 15
b = -2
c = 2Y
d = -1
My equation for the working out of the number of cubes in a 3D version of the pattern is:
_(nƒ) - 2n<sup>2</sup> + 2Y(n) - 1
Testing out the New Equation
I will take sequence 10 to try and test this new equation.
Sequence 5 :
N = 5
_(5ƒ) - 2(5<sup>2</sup>) + 2Y(5) - 1 =
166Y - 50 + 135 - 1 =
129
The formula on this sequence seems to be successful.
I will now apply it to another sequence to be 100% correct:
N = 4
_(4ƒ) - 2(4<sup>2</sup>) + 2Y(4) - 1 =
855 - 32 + 10Y - 1 =
63
The formula again proves to be successful.
Using the formula to find the number of squares in a higher sequuence
not yet explored in this investigation.
Sequence 10:
N = 10
_(10ƒ) - 2(10<sup>2</sup>) + 2Y(10) - 1 =
13335 - 200 + 26Y - 1 =
1159
Sequence 15:
N = 15
_(15ƒ) - 2(15<sup>2</sup>) + 2Y(15) - 1 =
4500 - 450 + 40 - 1 =
4089
This equation has correctly given me the number of squares in each
sequence which again proves it can be applied to any of the 3D sequences to give the correct answer.
My Conclusions
I have made a number of conclusions from the investigation I have carried out.
Firstly I have deciferred that the equation used in the 2D pattern
was a quadratic. This can be proven through the fact that the 2nd
difference was a constant, a necessary element of any quadratic
and also the fact that the first value has to be squared. This can
also be proved by illustrating the equation on the graph, creating a curve.
I have also established that the top triangular half of the 2D pattern
always turns out to be a square number.
If we now look at the 3D pattern, the equation I achieved for it
has turned out to a cubic equation. This can be proven through the
constant, again a necessary characteristic of any cubic equation
and also the fact that its 1st value must be cubed and its second
squared. If we drew a graph we would get a ccurved graph in which
the line falls steeply, levels off and then falls again.
The Differentiation Method developed by Jean Holderness played a
very important role in this investigation. It helped us to gain
knowledge of any pattern and anything that would help in the invetigation,
giving us our constant, but most importantly it gave us the equation
on which to base our solutions.
It was:
an<sup>2</sup> + bn + c
This proved very helpful.
To find our equation we then substituted in different values which
we could find in our differentiation table.
I have concluded that both the equations proved to be very successful.
Therefore the equations are:
For the 2D pattern the equation is;
2n<sup>2</sup> - 2n + 1
For the 3D pattern the equation is;
_(nƒ) - 2n<sup>2</sup> + 2Y(n) - 1