# Number Stairs

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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.

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 |

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

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