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  • Level: GCSE
  • Subject: Maths
  • Word count: 4693

Maths grid extension

Extracts from this document...

Introduction

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

                + 6image00.pngimage01.png

+ 6image04.pngimage01.png

+ 6image01.png

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 ‘Unis the term which is the stair total. I have realised that the ‘6n’ in the formula must come from the number of squares in the stair shape.

...read more.

Middle

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.

image02.png

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 ‘Unis the term which is the stair total – would also always be the same no matter where this stair shape is translated around any size grid.

As a check to support this theory, I will substitute ‘n’ for 11 on the same 5 x 5 grid.

21

22

23

24

25

16

17

18

19

20

11

12

13

14

15

6

7

8

9

10

1

2

3

4

5

Stair-number (n) = 11

The grid is still 5 x 5, so ‘g’ = 5.

Stair-total (Un)         = 6n + 4g + 4

(found using        = 6(11) + 4(5) + 4        

the formula)        = 66 + 20 + 4

                = 90

Stair-total (found by adding) = 11 + 12 + 13 + 16 + 17 + 21 = 90

This shows that my formula must work for all 3-stair shapes on the 5 x 5 grid.

To investigate if this formula also works for 3-stair shapes on other grid sizes, I am going to see if the formula still works for 3-stair shapes on a 10 x 10 grid.

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

Grid: 10 x 10

‘n’ = 52, ‘g’ = 10:

Stair-total (Un)        =         6n + 4g + 4

(found using=         6(52) + 4(10) + 4

the formula)        =         312 + 40 + 4

                =         356

Stair-total         =         52 + 53 + 54 + 62 + (found by                 63 + 72

adding)

=356

This must mean that my formula Un = 6n + 4g + 4 – where ‘n’ is the stair number, ‘g’ is the grid size, and ‘Unis the term which is the stair total – works for all 3-stair numbers on any size grid.

Something I have noticed is that because this formula Un = 6n + 4g + 4 works for 3-stair shapes on a 10 x 10 grid, it also confirms the formula Un = 6n + 44 for 3-stair shapes on 10 x 10 grids, which I had previously found.

g = 10 (because the grid is 10 x 10)

Un         = 6n + 4g + 4

= 6n + 4(10) + 4

= 6n + 40 + 4

= 6n + 44image02.png

Next I will find a formula for calculating the stair-total of a 4-stair shape for any given stair-number.

Stairs: 4

To find a formula for calculating the stair total for any given stair-number of a 4-stair shape on a     5 x 5 grid, I will draw out the 4-stair shape on a 5 x 5 grid in terms of ‘n’ and ‘g’.

21

22

23

24

25

16

n + 3g

18

19

20

11

n + 2g

n + 2g + 1

14

15

6

n + g

n + g + 1

n + g + 2

10

1

n

n + 1

n + 2

n + 3

Grid: 5 x 5

The stair total for this 3-stair shape is

n + (n + 1) + (n + 2) + (n + 3) + (n + g) + (n + g + 1) + (n + g + 2) + (n + 2g) + (n + 2g + 1) + (n + 3g)

= n + n + 1 + n + 2 + n + 3 + n + g + n + g + 1 + n + g + 2 + n + 2g + n + 2g + 1 + n + 3g

= 10n + 10g + 10.

I think that the formula Un = 10n + 10g + 10 for working out the stair total for any given stair number – where ‘n’ is the stair number, ‘g’ is the grid size, and ‘Unis the term which is the stair total – would also always be the same no matter where this stair shape is translated around any size grid.

As a check to support this theory, I will substitute ‘n’ for 6 on the same 5 x 5 grid.

21

22

23

24

25

16

17

18

19

20

11

12

13

14

15

6

7

8

9

10

1

2

3

4

5

Stair-number (n) = 6

The grid is still 5 x 5, so ‘g’ = 5.

Stair-total (Un)         = 10n + 10g + 10

(found using        = 10(6) + 10(5) + 10        

the formula)        = 60 + 50 + 10

                = 120

Stair-total (found by adding) = 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 16 + 17 + 21 = 120

This shows that my formula must work for all 4-stair shapes on the 5 x 5 grid.


To investigate if this formula also works for 4-stair shapes on other grid sizes, I am going to see if the formula still works for 4-stair shapes on a 7x 7 grid.

43

44

45

46

47

48

49

36

37

38

39

40

41

42

29

30

31

32

33

34

35

22

23

24

25

26

27

28

15

16

17

18

19

20

21

8

9

10

11

12

13

14

1

2

3

4

5

6

7

Grid: 7

‘n’ = 22, ‘g’ = 7:

Stair-total (Un)        =         10n + 10g + 10

(found using=         10(22) + 10(7) + 10

the formula)        =         220 + 70 + 10

                =         300

Stair-total         =         22 + 23 + 24 + 25 + 29 + 30 + 31 + 36              (found by                     37 + 43

adding)

=300

This must mean that my formula Un = 10n + 10g + 10 – where ‘n’ is the stair number, ‘g’ is the grid size, and ‘Unis the term which is the stair total – works for all 4-stair numbers on any size grid.

image02.png

Next I will find a formula for calculating the stair-total of a 5-stair shape for any given stair-number.

Stairs: 5

To find a formula for calculating the stair total

for any given stair-number of a 5-stair shape I will use an 8 x 8 grid because there might not be enough room to check a formula by translations later on, on a 5 x 5 grid. I will draw out the 5-stair shape on an 8 x 8 grid in terms of ‘n’ and ‘g’.

57

58

59

60

61

62

63

64

49

50

51

52

53

54

55

56

41

n + 4g

43

44

45

46

47

48

33

n + 3g

n + 3g + 1

36

37

38

39

40

25

n + 2g

n + 2g + 1

n + 2g + 2

29

30

31

32

17

n + g

n + g + 1

n + g + 2

n + g + 3

22

23

24

9

n

n + 1

n + 2

n + 3

n + 4

15

16

1

2

3

4

5

6

7

8

...read more.

Conclusion

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                --------

Term 4-term 3: 37a + 7b + c = 6 … this will become number 5

Term 3- term 2: 19a + 5b +c = 3… this will become number 6

Term 2- term1: 7a + 3b + c =1… This will become number 7

Term 5 – term 6: 18a + 2b = 3 … this will become number 8

Term6 – term 7: 12a + 2b = 2 …This will become number 9

Term 8- term 9: 6a = 1

A = 1/6

Therefore term 8 becomes 3 + 2b = 3

Which means that b = 0

In term 7: 7/6 + c = 1

Which means that C = -1/6

In term 1 A and C cancel each other out, and with B equalling zero it must mean that D also equals zero.

Therefore the expression is 1/6x3 – 1/6x

e.g. x = 3

27/6 – ½ = 4

This fits in with the above table, to make sure another number will be substituted

X = 5

125/6 – 5/6 = 120/6 = 20

Again this fits into the table meaning that my formula is correct.

...read more.

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