• 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
  12. 12
    12
  13. 13
    13
  14. 14
    14
  15. 15
    15
  16. 16
    16
  17. 17
    17
  18. 18
    18
  19. 19
    19
  20. 20
    20
  21. 21
    21
  • Level: GCSE
  • Subject: Maths
  • Word count: 4685

Maths coursework. For my extension piece I decided to investigate stairs that ascend along with the numbers, in order to do this the grid was turned upside-down. I aim to see if there is a pattern within these stairs

Extracts from this document...

Introduction

 For my extension piece I decided to investigate stairs that ascend along with the numbers, in order to do this the grid was turned upside-down. I aim to see if there is a pattern within these ‘stairs’

91

92

93

94

95

96

97

98

99

100

81

82

83

84

85

86

87

88

89

90

71

72

73

74

75

76

77

78

79

80

61

62

63

64

65

66

67

68

69

70

51

52

53

54

55

56

57

58

59

60

41

42

43

44

45

46

47

48

49

50

31

32

33

34

35

36

37

38

39

40

21

22

23

24

25

26

27

28

29

30

11

12

13

14

15

16

17

18

19

20

1

2

3

4

5

6

7

8

9

10

This is a 10 x 10 size grid with a 3-stair shape in blue. This is called the stair total. The stair total for this stair shape is 25 + 26 + 27 + 35 + 36 + 45 = 194. To investigate the relationship between the stair total and the position of the stair shape, I will use the far-left bottom square as my stair number:

 This is always the smallest number in the stair shape. It is 25 for this stair shape.

I will then translate this 3-stair shape to different positions around this 10 x 10 grid:

46

36

37

26

27

28

The stair-total for this stair shape is 26 + 27 + 28 + 36 + 37 + 46 = 200

87

77

78

67

68

69

The stair-total for this stair shape is 67 + 68 + 69 + 77 + 78 + 87 = 446

88

78

79

68

69

70

The stair-total for this stair shape is 68 + 69 + 70 + 78 + 79 + 88 = 452

23

13

14

3

4

5

The stair-total for this stair shape is 3 + 4 + 5 + 13 + 14 + 23 = 62

24

14

15

4

5

6

The stair-total for this stair shape is 4 + 5 + 6 + 14 + 15 + 23 = 68

Stair number

Stair Total

25

194

26

200

67

446

68

452

3

62

4

68

I will then summarize these results in a table:

In order to find a formula that I can use to find the stair total when I am given the stair number, I am going to put the stair number as the position and the stair total as the term for the sequence:

Position

25

26

67

68

3

4

Term

194

200

446

452

62

68

                + 6

+ 6

+ 6

I have noticed that there is an increase of 6 between two consecutive terms in this arithmetic sequence. Therefore the position-to-term rule must be 6n + or – something.

Position (n)

25

26

67

68

3

4

Term (u)

194

200

446

452

62

68

6n

150

156

402

408

18

24

+ 44

+ 44

+ 44

+ 44

+ 44

+ 44

As can be seen, the term is always 6 times the position, plus 44.

Thus, for the bottom stair of any 3-stair shape on a 10 x 10 grid, the formula must be Un = 6n + 44, where ‘n’ is the stair number, and ‘Un

...read more.

Middle

        = 21 + 5 + 1

                        = 27

Stair-total (found by adding) = 7 + 8 + 12 = 27

This shows that my formula must work for all 2-stair numbers on the 5 x 5 grid.

However I was yet to find out if this formula would also work on other grid sizes. To investigate this, I am going to see if the formula still works for 2-stair shapes on an 8 x 8 grid.

57

58

59

60

61

62

63

64

49

50

51

52

53

54

55

56

41

42

43

44

45

46

47

48

33

34

35

36

37

38

39

40

25

26

27

28

29

30

31

32

17

18

19

20

21

22

23

24

9

10

11

12

13

14

15

16

1

2

3

4

5

6

7

8

Grid: 8 x 8

‘n’ = 46, ‘g’ = 8:

Stair-total (Un)    =          3n + g + 1

(found using=          3(46) + (8) + 1

the formula)        =          138 + 9

                        =          147

Stair-total           =          46 + 47 + 54

(found by            =          147

adding)

This must mean that my formula Un = 3n + g + 1 – where ‘n’ is the stair number, ‘g’ is the grid size, and ‘Unis the term which is the stair total – works for all 2-stair shapes on any size grid.

Next I will find a formula for calculating the stair-total of a 3-stair shape for any given stair-number. So far I only know the formula for working out the stair-total of 3-stair shape for a 10 x 10 grid.

Stairs: 3

To find a formula for calculating the stair total for any given stair-number of a 3-stair shape on a  5 x 5 grid, I will draw out the 3-stair shape on a 5 x 5 grid in terms of ‘n’ and ‘g’.

Grid: 5 x 5

21

22

23

24

25

16

n + 2g

18

19

20

11

n + g

n + g + 1

14

15

6

n

n + 1

n + 2

10

1

2

3

4

5

The stair total for this 3-stair shape is

   n + (n + 1) + (n + 2) + (n + g) + (n + g + 1) + (n + 2g)

= n + n + 1 + n + 2 + n + g + n + g + 1 + n + 2g

= 6n + 4g + 4.

As said before, just as these individual values in each square would always be the same no matter where this 3-stair shape is translated around any size grid, the formula Un = 6n + 4g + 4 for working out the stair total for any given stair number – where ‘n’ is the stair number, ‘g’ is the grid size, and ‘Un

...read more.

Conclusion

I have noticed that all the formulae are made up of three terms: an ‘n’ term, a ‘g’ term and a ‘number’ term (the first formula could be written as 1n + 0g + 0). These terms are always added together to form the formula. The coefficient of ‘g’ is always the same as the number in the ‘number’ term.

Therefore I can tackle this problem of finding a rule connecting the number of stairs and its formula for working out the stair total in two parts: finding the rule connecting the number of stairs and the coefficient of the ‘n’ term, and then finding a rule connecting the number of stairs and the coefficient of the second (and in effect the third term as well).

Hence, firstly, I will write out the sequence where I am going to put the number of stairs as the position and the ‘n’ term as the term for the sequence:

position

1

2

3

4

5

6

term

1

3

6

10

15

21

I have realised that this sequence is made up of the triangle numbers. From previous work I know that the formula for term ‘t’ in the sequence is ½ t (t + 1).

Therefore the 7th term will be ½ (7) [(7) + 1] = ½ 7 (8) = ½  x 56 = 28

To check this I will add up the numbers that form a triangle with 7 rows:

1 + 2 + 3 + 4 + 5 + 6 + 7 = 28.

Hence, the formula for working out the stair total of a 7-stair shape will start like this: 28n + …


Position

1

2

3

4

5

6

Term

0

1

4

10

20

35

1st difference

+ 1

+ 3

+ 6

+ 10

+ 15

2nd difference

+ 2

+ 3

+ 4

+ 5

3rd difference

+ 1

+ 1

+ 1

As there is a 3rd difference in this sequence, this means that it is a cubic equation.

Therefore, if I use the cubic equation ax3+ bx2 + cx+ d, I will be able to find the 4 unknowns.

Term 1:             a(1) 3+ b(1) 2+ c(1)+ d = 0

a + b + c + d = 0                       --------

Term 2:             a(2)3+ b(2)2 + c(2)+ d = 1

8a + 4b + 2c + d = 1                  --------

Term 3:             a(3)3+ b(3)2 + c(3)+ d = 4

27a + 9b + 3c + d = 4                --------

Term 4:             a(4)3+ b(4)2 + c(4)+ d = 10

64a + 16b + 4c + d = 10                        --------

...read more.

This student written piece of work is one of many that can be found in our GCSE Number Stairs, Grids and Sequences 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 Number Stairs, Grids and Sequences essays

  1. Number stairs

    This means I have proved for my formula to be correct Now I am going to use the same logic and method for one other grid and that is a 12 by 12 Number Grid. I will use the same 3 step-stair approach and I can then use the algebra

  2. GCSE Maths Sequences Coursework

    Nth term = 3N+? From my table I can see that 3 times N gives me the number in the sequence so; Nth term for Shaded = 3N Stage (N) 1 2 3 Sequence 1 4 10 1st difference +3 +6 2nd difference +3 Unshaded I can see here that there is not much

  1. Number Grid Coursework

    79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 Fig 2.5 Fig 2.1 to Fig 2.5 are the grids used for the varying values of z.

  2. Number Grids Investigation Coursework

    (n - 1) I will now try this formula on a completely different example, a 4 x 5 rectangle on a grid with a width of 8 and differences between the numbers on the grid of 3. D = wp2 (m - 1)

  1. Algebra Investigation - Grid Square and Cube Relationships

    matter what number is chosen to begin with (n), a difference of 4ghw-4gh-4gw+4g will always be present. Testing: Using the experimental number box above, it is possible to prove that the formula works, and is correct. The box had a height and width of 2x2, and was based on a 5x5 grid, with an increment size (s)

  2. Maths - number grid

    - s(s+24) s(s+20)+4(s+20) - s -24s s +20s +4s+80 - s -24s =80 I have now calculated the answer for my 3x2 rectangles and came to a difference of 20 and an answer for my 5x3 rectangles and came to a defined difference of 80.

  1. Maths Grids Totals

    I am going to use a 13 x 13 grid with which to confirm the rule. 13 x 13 Grids I am first going to start with some 2 x 3 grids, then 3 x 4, and then 4 x 5.

  2. Maths-Number Grid

    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 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74

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