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Introduction

James Hooper        Maths Coursework        04/05/2007

Page

Maths Investigation – Pile ‘em High

“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          Middle 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=n² 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

Conclusion

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

ax² + bx + c

2x² + 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(2x2 – 1)

= 6

This is correct

3(2x3 – 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.

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

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