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“Jo has started in a local supermarket and her first job is to build displays of soup tins. To make them stable they are stacked using a brick bond so that each tins stands on two others. The tins are stacked against the wall. Each stack is complete with one tin in the top row.”

This piece of Maths coursework is about investigating sequences from a practical situation. In this investigation, tins are used to build stacks using a brick bond so that each new tin stands on two others. The tins are stacked flat against a structure and each stack is complete with one tin in the top row.

An Example:

First I will investigate a two row stack.

In this two row stack there are:

Two tins on the base (row 2)

One tin on the top (row 1)

Three tins altogether

In a three row stack there are:

Three tins on the base (row 3)

Two tins in the middle (row 2)

One tin on the top (row 1)

Six tins altogether

An early pattern that I can see is that whatever the row number (counting down from row 1 at the top) there are that many tins in the row e.g.

row 3 – 3 tins. It appears to me that there is some kind of pattern forming with the total number of tins in each built up stacks. (See table)

Number of Stacks

The Amount of Tins

2

3

3

6

4

10

5

?

I predict that for five stacks, the amount of tins needed will be fifteen based on other stacks e.g. for two stacks there are three and then for three stacks there are six so two tins are added. Then as you go down the table the tins adds on another i.e. +2, +3, +4. I have tested it practically using real tins and for five stacks there were fifteen tins. I recognise this pattern. It is the sequence of triangular numbers.

N= 1, 2, 3, 4, 5 (term number)

1 3 6 10 15 …….line 1

+2 +3 +4 +5 …….line 2

1 1 1 …….line 3

The numbers on line 1 are the sequence of total number of tins in each stack and those on the second line are between the numbers in the sequence.

I have found a formula for this sequence using rules of quadratic sequences. I did this by:

1, 2, 3, 4, 5 (term number)

a+b+c = 1 3 6 10 15 …….line 1

3a+b = +2 +3 +4 +5 …….line 2

2a = 1 1 1 …….line 3

ax� + bx + c is the standard quadratic sequence and is used to find the formula

So…. 2a= 1

a= 0.5

3a+b = 2

1.5+b =2

b=o.5

a+b+c=1

0.5+0.5+c=1

c=0

So by substituting the answers into the formula on the previous page, I will get an overall formula for this investigation.

ax� + bx + c

0.5x� + 0.5x + c

Therefore the formula I have found is when x = The number of tin rows:

tn = 1/2x� + 1/2x

I will now test this formula to make sure it works in all stacks.

Firstly I will try two:

0.5 x 2� + 0.5 x 2

I calculated this sum and the answer was three. This was correct. I tried it again but with four;

0.5 x 4� + 0.5 x 4 = 4

This was also correct. This means that the formula for the brick bonding was correct and my prediction was correct.

I am now able to answer question one which is to find how many tins are in a stack of nineteen tins high.

0.5 x n� + 0.5 x n

0.5 x 19� + 0.5 x 19

= 190

Therefore there are 190 tins in a stack of 19 tins high.

Now I am going to investigate a different structure to see if the formula is the same for a four tin base that has a tin on top that touches all four other tins.

An Example :

I will find a sequence for the number of tins used and will then investigate a formula for a square based tin construction.

Term

Number= 1 2 3

Base= 1 4 9

I have recognised that there is a pattern for the number of tins on the base of a four based stack. There is a repetition of square numbers i.e.

1 x 1 = 1

2 x 2 = 4

3 x 3 = 9

Therefore the formula for the base is tn=n� but I am trying to investigate the number of tins in the whole structure.

I have found a sequence by using a practical and counting each tin in each stack.

Term Number 1, 2, 3, 4, 5

Total Number of Tins 1, 5, 14, 30, 55

Square Number Differences 4 9 16 25

(Base numbers)

These are odds numbers +5 +7 +9

missing numbers 1 and 3

Constant +2 +2

The reason why row four is tw0 is because the difference between odd numbers is always two. As we have to come down to a third line we are now in a cubic situation. I now have to use the cubic standard sequence t0 find the formula for the number of tins

Term Number 1, 2, 3, 4, 5

N= 1, 5, 14, 30, 55

a+b+c= 4 9 16 25

7a+3b+= +5 +7 +9

12a+2b= +2 +2

The standard cubic formula is an� + bn� + cn + d

Therefore by substituting again, I can find the formula.

6a = 2

a =

12a + 2b + 5

4 + 2b = 5

b = 1/2

7a + 3b + c = 4

7 + 1 1/2 + c = 4

c =

+ 1/2 + = 1

d = 0

Consequently the formula for a square based tin structure is….

tn = n� + 1/2n� + n

n = Term Number

Further into my investigation, I will investigate a star based structure by adding on one more tin on to the end of each line every time. I will try and find a formula and test it to make sure it works. I will use a standard sequence rule and put numbers into it.

Firstly I will find a sequence to this structure that I am investigating. I will use the base numbers to as a starting point to investigating a formula

Term Number

1 2 3

Number of Tins on Base

1 5 9

Tins Added Each Time

1 6 15

I tested this in a practical situation and found that for term number four there were 28 tins altogether.

I will now use the numbers to make a sequence;

Term Number (N) = 1, 2, 3, 4

Added Tins 1, 6, 15, 28

Difference 5 9 13

Constant +4 +4

This shows that I will have to use the quadratic formula as there are three rows of working in the sequence above.

1, 2, 3, 4, (term number)

a+b+c = 1 6, 15, 28 …….line 1

3a+b = 5 9 13 …….line 2

2a = +4 +4 …….line 3

Therefore;

2a= 4

a= 2

3a+b = 5

6+b =5

b=-1

a+b+c=1

2-1+c=1

c=0

ax� + bx + c

2x� +- 1x which can be simplified into tn = n (2n – 1)

Therefore the formula for this star based structure is tn = n (2n – 1)

I will now test this formula to check that this formula works in all cases:

2(2×2 – 1)

= 6

This is correct

3(2×3 – 1)

= 15

This is also correct

This means that the formula that I have investigated is correct for this particular pattern, where x = the term number.

Conclusion

1

1

2

5

3

14

4

30

5

55

6

91

7

140

8

204

9

285

10

385

1

1

2

3

3

6

4

10

5

15

6

21

7

28

8

36

9

45

10

55

1

1

2

6

3

15

4

28

5

45

6

66

7

91

8

120

9

153

10

190

This graph shows all the patterns onto one graph. It shows that a star based structure uses more tin cans than a brick bond. They all have a positive relationship as all the lines move upwards. The star based structure and square based structure have the same amount of tins until about row four and then they start to split of and the square based tin stacks stack to decrease compared to the star based structure. A brick bond uses the smallest amount of tins overall.

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