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• Level: GCSE
• Subject: Maths
• Word count: 1752

# Border coursework

Extracts from this document...

Introduction

Borders Courseworkcoec ec" . "r se" . ec . "ec" . "w or". ec . " " . ec . "k inec foec " . ec . ".

Aim: To investigate the sequence of squares in a pattern as shown below:

In this investigation, I have been asked to find out how many squares would be needed to make up a certain pattern according to its sequence. In this investigation I hope to find a formula which could be used to find out the number of squares needed to build the pattern at any sequential position. Firstly I will break the problem down into simple steps to begin with and go into more detail to explain my solutions such as the nth term. I will illustrate fully any methods I should use and explain how I applied them to this certain problem. I will firstly carry out this experiment on
a 2D pattern and then extend my investigation to 3D.
coca ca" . "r se" . ca . "ca" . "w or". ca . " " . ca . "k inca foca " . ca . "!

Apparatus:

Variety of sources of information

A calculator
A pencil
A pen
Paper
Ruler

A computer to work out equations on

I have come up with the following numbers and sequences. This was done by drawing out the sequence.cofd fd" . "r se" . fd . "fd" . "w or". fd . " " . fd . "k infd fofd " . fd . ";

 Seq no 1 2 3 4 5 6 7 No of squares 1 5 13 25 41 61 85

Middle

32

50

72

2n^2 + ?

1

5

13

25

41

61

2n^2 +2n+1

1

5

13

25

41

61

I can also form an equation through the trial and improvement method.
1) 2(n -1) (n - 1) + 2n - 1

2) 2(n
2 - 2n + 1) + 2n - 1

3) 2n
2 - 4n + 2 + 2n - 1

4) 2n
2 - 2n + 1

Therefore my final equation is:
2n
2 - 2n + 1

Proving My Equation and Using it to Find the Number
of Squares in Higher Sequences

I can deduct from this table and the formulas that the nth term formula can be 2n2 + 2n + 1. There is one problem with this formula. If we take the first sequence to be one square, the formula will not work. Therefore to allow the first term to be 1, I will use the formula 2n2 - 2n + 1. This formula does not change any of the other sequence results.

I will now prove my equation in a number of sequences, including higher sequences that I yet have to explore.

Sequence 3:
1. 2(3
2) - 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
2) - 10 + 1
2. 2(25) - 10 + 1
3. 50 - 10 + 1
4. 50 - 9
5. = 41
Successful

Sequence 6:
1. 2(6
2) - 12 + 1
2. 2(36) - 12 +1
3. 72 - 12 + 1
4. 72 - 11
5. = 61
Successful

Sequence 8:
1. 2(8
2) - 16 + 1
2. 2(64) - 16 + 1
3. 128 - 16 + 1
4. 128 - 15 = 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
2 - 2n + 1 to try and find
the number of squares contained in sequence 20.

Sequence 20:

2 (20
2) - 40 + 1
2(400) - 40 + 1
800 - 40 + 1
800 - 39
= 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
I used 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 for a 2d sequence.

3D sequence

The sequence for the 3d sequence is as follows. I built the pattern to get the no of blocks in the first few patterns.

 n 1 2 3 4 No of blocks 1 7 63 129

Now I will try to establish the number of squares in a 3d pattern. As stated before the equation for establishing the 2d pattern is:

2n2 -2n+1

This gives the pattern:

 Seq no 2 3 4 5 6 7 No of squares 5 13 25 41 61 81

Conclusion

(1+5+13+25+41+61+41+25+13+5+1)= 231.

This therefore proves that my sequence is correct , and thus proved.

Conclusion

I have made a number of conclusions from the investigation I have carried out.

Firstly I have found out 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, which is a necessary element of any quadratic. Also as the 2nd difference between the no of blocks was 4, Therefore I knew that the quadratic equation had to include a 2 in it. This allowed me to find out the 2D nth term formula

For the 3D equation, I have discovered that the pattern follows a symmetrical triangle consisting of the numbers from 2D. Thus by using the formula 2n2 -2n+1 from the 2D I have been able to use summation to work out the value of the nth number. The summation function allowed me to sum the results of [2n2 -2n+1] from n=1 to n= n-1. Using this result and combining it with the value of [2n2 -2n+1] at n, I have worked out the number of block squares.

a51m. Thus, we can say that whilst this represents a progression, in the end we have come no closer to any "real" knowledge.

this coursework was downloaded from Coursework.info. Redistribution prohibited. © 2004-2006 Student Media Services Limited. Redistribution and retransmission explicitly prohibited.

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