• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
• Level: GCSE
• Subject: Maths
• Word count: 1642

# Investigate the sequence of squares in a pattern.

Extracts from this document...

Introduction

Yasotharan Paramesparan        Maths coursework

Borders Coursework

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.

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.

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

I will use these numbers to try to create a type of formula to get any no of squares in any sequence.

 1 1+3+1 5 2 1+3+5+3+1 13 3 1+3+5+7+5+3+1 25 4 1+3+5+7+9+7+5+3+1 41 5 1+3+5+7+9+11+9+7+5+3+1 61 6 1+3+5+7+9+11+13+11+9+7+5+3+1 85

Middle

2n^2

2

8

18

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.

10 RP        Mr O’Connell

This student written piece of work is one of many that can be found in our GCSE Consecutive Numbers section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related GCSE Consecutive Numbers essays

1. ## GCSE Maths Coursework - Maxi Product

I will try now in fractional numbers if I can get a number higher than 64. (8 1/10,7 9/10)= 16 --> 8 1/10+7 9/10 --> 8 1/10x7 9/10 =63.99 (8 4/15,7 11/15)= 16 --> 8 4/15+7 11/15 --> 8 4/15x7 11/15=63.93 (2dp)

2. ## Consecutive Numbers Investigation

20.12 = 20.1*20.1 = 404.01 Difference 144.8 It would appear that it works with decimals. I will now try negative numbers. -6, -10 -62 = -6*-6 = 36 -102 = -10*-10 = 100 Difference 64 -42, -46 -422 = -42*-42 = 1764 -462 = -46*-46 = 2116 Difference 352 -23,

1. ## Chessboard coursework

N th rule= nx6+2 =6n+2 The reason in this that you don't get square numbers as answers is because the shapes are not square. Extension 2 The cubic formula of type 3 2 Y= Ax + Bx + Cx + D 1. A+B+C+D=2 2. 8A+4B+2C+D=8 3.

2. ## I am to conduct an investigation involving a number grid.

(x + 20) x2 + 20x + x + 20 = (x2 + 23x + 20)

1. ## Fraction Differences

And substituted the n for 1: 1(1 + 1)(1 + 2)(1 + 3) = 24 The answer I had was 24. The sequence that I was working from gave me 4 as the first number, this would imply that I would have to divide by 6 to get my first

2. ## Borders and squares

but since here we want to find a formula with the nth term we will write it in a form of (2n2) Then times each of our sequence numbers by it, and then add it to No. Of cubes. 'n' indicates the position in the sequence.

1. ## Analyse the title sequences of two TV programmes, comparing and contrasting the techniques used ...

The 'NYPD Blue' title sequence transports the viewer to the wild streets of New York as we are immediately thrown into the depths of the urban jungle. However, 'The Bill' is obviously set in London even though the viewer is not shown any famous landmarks.

2. ## Investigating a Sequence of Numbers.

+ 2 x 2! + 3 x 3! = 1 + 4 + 18 = 23 S4 = a1 + a2 + a3 + a4 = 1 x 1! + 2 x 2! + 3 x 3! + 4 x 4! = 1 + 4 + 18 + 96 = 119 S7 = a1

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to