# Staircase Coursework

Extracts from this document...

Introduction

Dominik Borgolte

Math GCSE Coursework

Dominik Borgolte

Buckswood School

GCSE Mathematic coursework

“Step stairs”

Centre number:

Candidate number: 2776

## Introduction

I am going to investigate formulas that can be used to calculate the sum of the number in a staircase on a grid paper.

In my investigation I will use:

St: stand for stair total ( S stand for stair and t for the total sum )

g: stand for the size of the grid (e.g. in 12x12 grid n=12)

n: stand for stair number, which is always the bottom left number

x. stand for the size of the stairs ( e.g. a 6 step stair would be x=6 )

## 3 step stair on a 10x10 grid

## ( 1 )

I want to find a formula, which makes it easier for me to calculate the total size of a 3 step stair in a 10 x 10 grid.

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 | 31 | 32 | 33 | 34 | 35 | 36 |

41 | 42 | 43 | 44 | 45 | 46 | 47 | 48 | 49 | 50 | 25 | 26 | 27 | 28 | 29 | 30 |

31 | 32 | 33 | 34 | 35 | 36 | 37 | 38 | 39 | 40 | 19 | 20 | 21 | 22 | 23 | 24 |

21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 | 13 | 14 | 15 | 16 | 17 | 18 |

11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 7 | 8 | 9 | 10 | 11 | 12 |

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 1 | 2 | 3 | 4 | 5 | 6 |

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

I have added all the squares of the 3 step stairs together, which gives me a result of 50, for step1.

I will now find a formula for the total, which will make it easier to calculate the sum.

I will label a stair algebraically

This is an example of stair 1

n stands for the number in the left bottom corner of the step.

In order of that if : 1 = n 2 = n + 1 and so on ................

n+20 | ||

n+10 | n+11 | |

n | n+1 | n+2 |

21 | ||

11 | 12 | |

1 | 2 | 3 |

=

Middle

177

178

179

180

151

152

153

154

155

256

257

258

259

160

161

162

163

164

165

136

137

138

139

140

141

142

143

144

145

146

147

148

149

150

121

122

123

124

125

126

127

128

129

130

131

132

133

134

135

106

107

108

109

110

111

112

113

114

115

116

117

118

119

120

91

92

93

94

95

96

97

98

99

100

101

102

103

104

105

76

77

78

79

80

81

82

83

84

85

86

87

88

89

90

61

62

63

64

65

66

67

68

69

70

71

72

73

74

75

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Formula: 6n + 4g +4

g = 15

n = 97

6x97 + 4x15 + 4 = 646

Test of the formula:

St: 97+98+99+112+113+127=646

So the formula also works for any other size grid

( 3a )

4 step stair on a 10x10 grid

## I will now try to find a formula for a 4 step stair on a 10 x 10 grid

4 step stair on a 10x10 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 |

I use the same principle like I already did with the 3 step stair on a 10x10 grid.

So I make the bottom left number n, which gives me a huge formula.

In this case

n + (n + 1) + (n + 2) + (n + 3) + (n + 10) + (n + 11) + (n + 12) + (n + 20) + (n + 21) +(n+30)

Then I add up all normal numbers and all the n`s

So in this case it would be 10 times n and all the numbers added up would be 110.

This leaves me with the final formula

St = 10n + 110

I will now test the formula to see if it is correct by using step 46

Because n= 46St = 10x46 + 110= 570

St = 46+47+48+49+56+57+58+66+67+76 = 570

So the formula is correct

## (3b) 4 step stair in a variable grid

## Now I will try to find a formula for a 4 step stair on any grid.

I will use a 15x15 grid for my investigation

4-step stair on 15x15 grid

46 | 47 | 48 | 49 | 50 | 51 | 52 | 53 | 54 | 55 | 56 | 57 | 58 | 59 | 60 |

31 | 32 | 33 | 34 | 35 | 36 | 37 | 38 | 39 | 40 | 41 | 42 | 43 | 44 | 45 |

16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 |

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |

Conclusion

But still a few improvements could be made. I could for example increase the amount of tests on different sizes of stairs and grids

I would try out a 5 or 8 step stair, or test them at different size grids like I already used, but in the end the final result, which is the overall formula

[1/2x (x+1)] n + (g+1) [1/6(x^{[4]}3-x)] would still be the same.

I had some help from my teacher and an older classmate. I also took one formula for the tetrahedral number out of the Internet. Other than that I did everything on my own, which is highly important in my opinion.

Therefore I think overall the coursework was a clear success, because I was able to find formulas for every different size step, and every different size grid investigation I made.

In the end I was even able to find a formula that can be used for any size grid and any size stair, so I could even achieve my final goal of the coursework.

I also think that this coursework was useful for my personal abilities, because I really improved my ability to think logically, and my ability to help my self when I am sucked at a certain point, which was the case (a few times) during this coursework.

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

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