• 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
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  • Level: GCSE
  • Subject: Maths
  • Word count: 1453

Maths Investigation - Pile 'em High

Extracts from this document...

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.

image01.pngimage01.pngimage01.pngimage01.pngimage01.pngimage00.pngimage01.png

An Example:

First I will investigate a two row stack.

image01.png

In this two row stack there are:

image01.pngimage01.png

Two tins on the base (row 2)

One tin on the top (row 1)

Three tins altogether

In a three row stack there are:image01.png

image01.pngimage01.pngimage01.pngimage01.pngimage01.png

Three tins on the base (row 3)

Two tins in the middle (row 2)

One tin on the top (row 1)

Six tins altogether

image01.png

image01.pngimage01.png

image01.pngimage01.pngimage01.png

image01.pngimage01.pngimage01.png

image01.png

...read more.

Middle

image02.png

An Example :         image02.pngimage02.png

image02.pngimage02.png

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

image04.pngimage04.pngimage04.pngimage04.pngimage04.pngimage04.pngimage04.pngimage04.pngimage04.pngimage04.pngimage04.pngimage04.png

image04.pngimage04.png

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    

...read more.

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

image08.png

image11.pngimage09.pngimage10.png

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.

...read more.

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

See related essaysSee related essays

Related GCSE Consecutive Numbers essays

  1. GCSE Maths Coursework - Maxi Product

    (7.1,6.9)= 14 --> 7.1+6.9 --> 7.1x6.9 =48.99 (7.2,6.8)= 14 --> 7.2+6.8 --> 7.2x6.8 =48.96 (7.01, 6.99)= 14 --> 7.01+6.99 --> 7.01x6.99=48.9999 I still have not yet found a number higher than 49 in decimal numbers. I will try now in fractional numbers if I can get a number higher than 49.

  2. In this investigation I will explore the relationship between a series of straight, non-parallel, ...

    In this diagram there are: Number of Lines (n) Cross-Over Points Open Regions Closed Regions Total Regions 1 0 2 0 2 2 1 4 0 4 3 3 6 1 7 4 6 8 3 11 5 10 10 6 16 Diagram 6: (all 6 lines cross each other,

  1. Nth Term Investigation

    The nth term is n x t / n x 2. For the fourth column (+) the numbers go up in 1's because when the length increase by one an extra + is put in the middle of the rectangle.

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

    X+11 X+12 X+13 X+4 26 27 28 29 30 X+20 X+21 X+22 X+23 X+24 36 37 38 39 40 X+30 X+31 X+32 X+33 X+34 46 47 48 49 50 X+40 X+41 X+42 X+43 X+44 [image015.gif] [image016.gif] 6 x 50 = 300 x (x + 44)

  1. Fraction Differences

    Fourth Sequence As I had hit on a vague pattern with the formulas I decided to investigate further by working out the differences between the last sequence to produce a new one: And then I applied the same formula pattern as before: n(n + 1)(n + 2)(n + 3)

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

    We glimpse a policewoman answering calls, symbolising that 'The Bill' will focus on real life issues in the day of the Metropolitan Police.

  1. Investigating a Sequence of Numbers.

    T4 = c1 + c2 + c3 + c4 = 5 + 22 + 114 + 696 = 837 T5 = c1 + c2 + c3 + c4 + c5 = 5 + 22 + 114 + 696 + 4920 = 5757 T6 = c1 + c2 + c3 +

  2. Number Grid Investigation.

    1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

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