# Number stairs.

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Introduction

Introduction

Below you will find a 10 by 10, 9 by 9, 8 by 8, 7 by 7 and 6 by 6 number grids all of whom who will show 5 3 step stairs. For each 3 step stair I will workout the total of them and then will go on to find where appropriate algebraic formulas. Once I have done this I will go on to see whether or not the formula which I have worked out for a 10X10 number will work on another sized grid and one all the grids have been completed I will create a table of results looking at the formulas which I have found and on what sized grid they belong to. It will be best to display my results on a table as it will be easier for me to read off my formulas.

Part 1

For part 1 will investigate the relationship between the stair total of the 3 step stairs and the position of the stairs on the number grid.

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 |

Stair 1:

73 + 74 + 75 + 83 + 84 + 93 = 482

A+A+1+A+2+A+10+A+10+1+A+20=

6A+44=482

6A=482-44

6A=438÷77=6

A=73

6(73) +44

Stair 2:

77 + 78 + 79 + 87 + 88 + 97 = 506

X+X+1+X+2+X+10+X+10+1+X+20 =

6X+44=506

6X=506-44

6X=462÷77=6

X=77

6(77) +44

Stair 3:

21 + 22 + 23 + 31 + 32 + 41 = 170

Y+Y+1+Y+2+Y+10+Y+10+1+Y+20=

6Y+44=170

6Y=170-44

6Y=126÷21=6

Y=21

6(21) +44

Stair 4:

25 + 26 + 27 + 35 + 36 + 45 = 194

L+L+1+L+2+L+10+L+10+1+L+20=

6L+44=194

6L=194-44=150

6L=150÷25=6

L=25

6(25) +44

Conclusion:

Based on all of the 4 3 step stairs that I have investigated, I can now say that I have found a formula which works with all of the other 3 step stairs. The formula which I have found is 6A+44 which just simply means 6AX4X10+4.

Middle

Grid Size: | Algebra: | General Formula: |

10X10 | 6Y+44 – stair 1 | 6Y X 4 X 10 + 4 |

9X9 | 6B+40 – stair 1 | 6B X 4 X 9 + 4 |

8X8 | 6F+36 – stair 1 | 6F X 4 X 8 + 4 |

7X7 | 6J+32 – stair 1 | 6J X 4 X 7 + 4 |

6X6 | 6N+28 – stair 1 | 6N X 4 X 6 + 4 |

From my table of results on the previous page, I have noticeda general trend:

I have noticed that as that grid size depreciates e.g. by one size from a 10X10 to a 9X9, the algebra also decreases by 4. This is because the grid size is getting smaller and therefore the numbers inside the grid are being reduced from 100 to 81.

13 | 14 | 15 | 16 |

9 | 10 | 11 | 12 |

5 | 6 | 7 | 8 |

1 | 2 | 3 | 4 |

- I will now test my theory that the size of the grid will depend the outcome of the algebra and the algebra will be 4 less then its predecessor e.g. the algebra changing from 6n + 28 on a 6X6 grid to 6n + 20 on a 4X4 grid. I will prove this by drawing up a 4X4 grid. The ‘Letter’ value in the grid is ‘n’.

1 + 2 + 3 + 5 + 6 + 9 = 26

n+n+1+n+2+n+4+n+4+1+n+8=

6n+20=26

6n=26-20=6

6n=6÷1=6

n=1

6(1) + 20

My test worked. It shows that the size of the grid always shows the outcome of the algebra and most importantly I have noticed the trend which has reduced to 20 as I predicted.

Extending my Original line of Enquiry:

Now that I have finished my 3 step stairs on different grid sizes, I will now extend my line of enquiry by investigating the

Conclusion

The formula for triangle numbers is:

nth term = ½ n (n+1), this is the same as my formula when ‘n’ and ‘x’ mean the same thing:

½ X (x+1)

Testing my formula:

½ x² + ½ x = ½ X (x+1)

when x = 4, ½ + 16 + ½ X 4 =10

8 + 2 = 10

Pyramid Numbers

The 3rd differences are the same, so as a result it is cubic. The cubic formula is:

a X³ + b X ² + c X + D

Substitute:

1x = 1: A + B + C + D = 0

2x = 2: 8A + 4B + 2D + D = 1

3x = 3: 27A + 9B + 3C + D =4

4x = 4: 64A + 16B + 4C + D = 10

8A + 4B + 2C + D = 1 64A + 16B + 4C + D = 10

A + B + C + D = 0 27A + 9B + 3C + D = 4

7A + 3B + c = 1 37A + 7B + C = 6

eliminates D: 27A + 9B + 3C + D = 4

8A + 4B + 2C + D = 1

19A + 5B + C = 3

19A + 5B + C = 3 37A + 7B + C = 6

7A + 3B + C = 1 eliminates C 19A + 5B + C = 3

12A + 2B = 2 18A + 2B = 3

to get B:

18A + 2B = 3 substitute: 12A + 2B = 2

12A + 2B = 2 a = ¹/6 12x ¹/6 + B = 2

6A = 1 12x ¹/6 + 0 = 2

A = ¹/6 B = 0

to get C: now we put our A, B + C values

into the cubic number formula:

Substitute: 7A + 3b + C = 1 a = ¹/6: a X ³ + b + x² + CX + d

A = ¹/6 7x ¹/6 + 3x D + C = 1 b = 0 : ¹/6 x³ + - ¹/6X + 0

B = ¹/6 1¹/6 + D + C = 1 c = - ¹/6: ¹/6 X³ + - ¹/6 X

d = 0

13 | 14 | 15 | 16 |

9 | 10 | 11 | 12 |

5 | 6 | 7 | 8 |

1 | 2 | 3 | 4 |

Now I am going to test my findings on my general rule, which is:

(½ ² + ½ ) n + (¹/6 X² = ¹/6x) g+ (¹/6 X² - ¹/6X) + (½ X 4² + ½ 4) + ¼ X 4 = 60

x = 4

n = 1

g = 4

My general rule for any grid size with any number of stairs has been proven to work.

Why you get triangle numbers and Why you get pyramid numbers:

1 Step Stair = 1n, 2 Step Stair = 3n + g + 1, 3 Step Stair = 6n + 6g + 4, 4 Step Stair = 10n +10g + 10

Candidate Name: Amish Patel Candidate Number: 8153 Centre Number: 12504

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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