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  • Level: GCSE
  • Subject: Maths
  • Word count: 3207

Number Stairs

Extracts from this document...

Introduction

Introduction:

The topic which I have been given for my GCSE mathematics coursework, is “Number Stairs”. Therefore this leads mainly to the two topics “sequences” and “algebra”.

In this investigation I am going to examine the number steps in different positions, .  For part one of my assignment I will use a 10 – by –10 grid and investigate the effect of different size number steps.  I will find out the nth term formula for the number steps.  I will solve this using both algebraically and numerically methods. I will refer to the number in the bottom left hand corner.

Then going on for the second part of my assignment I will use different size number grids and also use different shapes e.g. T shape. The number which I refer to for my nth term will depend on the kind of shape that it is.

 For this first part of my coursework I will be using a 10 – by – 10 grid, which is given to investigate the number steps to find the formula for different number steps.  My grid is shown below:

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

2 step stair:

I will firstly go about investigating the 2-step stair numerically, but before I do that I need to find the total of  at least seven steps.

...read more.

Middle

75

76

75+76+85 = 236

I have tested position 75 in a 10 –by – 10 grid, on a  2 step stair. After testing this I can therefore say that my formula has worked. So this shows me that my calculations which I have carried out are accurate.

Now I am  going to test the position 47 on a two step stair in a

10 – by – 10 grid.

Formula =3n+11

               = 3 x 47 +11

               = 152

Now I will see if the total number step adds up to 152. This will only work if the formula is accurate.

57

47

48

47+48+57 = 152

I have tested position 47 in a 10 –by – 10 grid, on a  2 step stair. After testing this I can therefore say that my formula has worked. So this shows me that my calculations which I have carried out are accurate.

I have tested out three different numbers within the 10 – by – 10 number grid on a 2 step stair and therefore I can now say that  my formula has worked out throughout working out the calculations for the 2 step stair. For that reason I would say my formula is accurate.

3-step stair:

I will now go about investigating the 3-step stair on a 10 –by – 10 grids. I will firstly investigate this numerically. I’m going to start with the number one and I will move one space to the right.  I am then going to find the nth term formula.

Position One :

21

11

12

1

2

3

Total of position 1:

1+2+3+11+12+21 = 50

Position Two :

22

12

13

2

3

4

Total of position 2:

2+3+4+12+13+22 = 56

Position Three :

23

13

14

3

4

5

Total of position 3:

3+4+5+13+14+23 =62

Position Four :

24

14

15

4

5

6

Total of position 4:

4+5+6+14+15+24 = 68

Position Five :

25

15

16

5

6

7

Total of position 5:

5+6+7+15+16+25 = 74

Position Six :

26

16

17

6

7

8

Total of position 6:

6+7+8+16+17+26 = 80

Position Seven :

27

17

18

7

8

9

Total of position 7:

7+8+9+17+18+27 = 86

Now that I have found out the totals of the first seven steps, I will be able to find the formula numerically for the 3-step stair:

nth term:

Position number:

1

2

3

4

5

6

7

Position Total:

50

56

62

68

74

80

86

1st Difference:

+6

+6

+6

+6

+6

nth term:

6n+44

Rule          = +6

nth Term    = 6n+b

1st term  50= 6(1) +b

               50= 6+b

                b= 50-6

                b= 44

nth term      = 6n+44

Now I have worked out the formula numerically and also by using the value of position 1 and also by finding the difference. I have both times got the same formula 6n+44. Now I will see if this gives me the value of position 3.

Formula = 6n+44

Position 5 = 6 x 3+44

This is correct as in the total for position 3, which I worked out previously and I got the answer 26. This shows me that I have got the right formula as it works.

I’m now going to find out the formula algebraically and see if I get the same formula. If they both mach then it is accurate.

n+20

n+10

n+11

n

n+1

n+2

n+n+n+n+n+n =6

1+2+10+11+20 = 44

End formula = 6n+44

After finding my formula algebraically as well I can see that the formula should be accurate. However just to make sure I will test the formula using 3 different numbers.

3-step stair: Testing

By using my formula 6n+44 I’m going to test this formula in three different positions referring to the number in the left hand corner.

I am firstly going to test the position 48 on a two step stair in a

10 –by – 10 grids:

Formula =6n+44

               = 3 x 48+11

               = 332

Now I will see if the total number step adds up to 332. This will only work if the formula is correct.

68

58

59

48

49

50

48+49+50+58+59+68 = 332

I have tested position 48 in a 10 –by – 10 grid, on a  3 step stair. After testing this I can therefore say that my formula has worked. So this shows me that my calculations, which I have carried out, are accurate.

Now going to test the position 22 on a two step stair in a

10 –by – 10 grids:

Formula =6n+44

               = 6 x 22 +44

               = 176

Now I will see if the total number step adds up to 176. This will only work if the formula is correct.

42

32

33

22

23

24

22+23+32+33+42 = 176

I have tested position 22 in a 10 –by – 10 grids, on a 3-step stair. After testing this I can therefore say that my formula has worked. So this shows me that my calculations that I have carried out are accurate.

Now I am going to test the position 58 on a two step stair in a

10 –by – 10 grids:

Formula =6n+44

               = 6x 58 +44

               = 392

Now I will see if the total number step adds up to 392. This will only work if the formula is accurate.

78

68

69

58

59

60

58+59+60+68+69+78 = 392

I have tested position 58 in a 10 –by – 10 grids, on a 3-step stair. After testing this I can therefore say that my formula has worked. So this shows me that my calculations that I have carried out are accurate.

I have tested out three different numbers within the 10 – by – 10 number grid on a 3 step stair and therefore I can now say that my formula has worked out throughout working out the calculations for the 3 step stair. For that reason I would say my formula is accurate.

4-step stair:

I will now go about investigating the 4-step stair on a 10 –by – 10 grids. I will firstly investigate this numerically. I’m going to start with the number one and I will move one space to the right.  I am then going to find the nth term formula.

Position One :

31

21

22

11

12

13

1

2

3

4

Total of position 1:

1+2+3+4+11+12+13+21+22+31 = 120

Position Two :

32

22

23

12

13

14

2

3

4

5


Total of position 2:

2+3+4+5+12+13+14+22+23+32 = 130

Position Three :

33

23

24

13

14

15

3

4

5

6

Total of position 3:

3+4+5+6+13+14+15+23+24+33 = 140

Position Four :

34

24

25

14

15

16

4

5

6

7

...read more.

Conclusion

I have tested out three different numbers within the 10 – by – 10 number grid on a 3 step stair and therefore I can now say that  my formula has worked out throughout working out the calculations for the 3 step stair. For that reason I would say my formula is accurate.

For this second part of my coursework I will be changing the sizes of the number grids and the shapes.  Firstly I am going to use the same

10 – by –10 grid and change the shapes.

Firstly I am going to use a C shape to investigate this formula.

Position One :

21

22

23

11

1

2

3

Total of position 1:

1+2+3+11+21+22+23 = 83

Position Two :

21

22

23

11

2

3

4

Total of position 2:

2+3+4+11+21+22+23 = 86

Position Three :

23

24

25

13

3

4

5

Total of position 3:

3+4+5+13+23+24+25 = 97

Position Four :

24

25

26

14

4

5

6

Total of position 4:

4+5+6+14+24+25+26 =

Position Five :

25

26

27

15

5

6

7

Total of position 5:

5+6+7+15+25+26+27 = 111

Position Six :

26

27

28

16

6

7

8

Total of position 6:

6+7+8+16+26+27+28 = 118

Position Seven :

27

28

29

17

7

8

9

Total of position 7:

7+8+9+17+27+28+29 = 125

Now that I have found out the totals of the first seven steps, I will be able to find the formula numerically for the 4-step stair:

nth term:

Position number:

1

2

3

4

5

6

7

Position Total:

83

86

97

104

111

118

125

1st Difference:

+7

+7

+7

+7

+7

nth term:

7n+76

Rule             = +7

nth Term       = 7n+b

1st term  83= 7(1) +b

               83= 7+b

                b= 83-7

                b= 76

nth term      = 7+76

Now I have worked out the formula numerically and also by using the value of position 1 and also by finding the difference. I have both times

Got the same formula 10n+110. Now I will see if this gives me the value of position 6.

Formula = 10n+110

Position 5 = 10 x 6+110 = 170

...read more.

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