# Sequences and series investigation

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Introduction

Sequences and series investigation By Neil

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.

The pattern is shown on the front page. 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. 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.

The Number of Squares in Each Sequence

I have achieved the following information by drawing out the pattern and extending upon it.

Seq. no. 1 2 3 4 5 6 7 8

No. Of cubes 1 5 13 25 41 61 85 113

I am going to use this next method to see if I can work out some sort of pattern:

Sequence Calculations Answer

1 =1 1

2 2(1)+3 5

3 2(1+3)+5 13

4 2(1+3+5)+7 25

5 2(1+3+5+7)+9 41

6 2(1+3+5+7+9)+11 61

7 2(1+3+5+7+9+11)+13 85

8 2(1+3+5+7+9+11+13)+15 113

9 2(1+3+5+7+9+11+13+15) +17 145

What I am doing above is shown with the aid of a diagram below;

If we take sequence 3:

2(1+3)+5=13

2(1 squares)

2(3 squares)

1(5 squares)

The Patterns I Have Noticied in Carrying Out the Previous Method

I have now carried out ny first investigation into the pattern and

have seen a number of different patterns.

Firstly I can see that the number of squares in each pattern is an odd number.

Middle

3. 18 -5

4. = 13

The formula when applied to sequence 3 appears to be

successful.

Sequence 5:

1. 2(52) - 10 + 1

2. 2(25) - 10 + 1

3. 50 - 10 + 1

4. 50 - 9

5. = 41

Successful

Sequence 6:

1. 2(62) - 12 + 1

2. 2(36) - 12 +1

3. 72 - 12 + 1

4. 72 - 11

5. = 61

Successful

Sequence 8:

1. 2(82) - 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 2n2 - 2n + 1 to try and find

the number of squares contained in sequence 20.

Sequence 20:

2 (202) - 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 + bn2 + 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

Conclusion

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:

an2 + 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;

2n2 - 2n + 1

For the 3D pattern the equation is;

_(nƒ) - 2n2 + 2Y(n) - 1

This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.

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