# Maths grid extension

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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 |

+ 6 | + 6 | + 6 |

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 ‘Un’is 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.

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 ‘Un’is the term which is the stair total – works for all 2-stair shapes on any size grid.

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 ‘Un’is 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 ‘Un’is 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 + 44

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 ‘Un’is 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 ‘Un’is the term which is the stair total – works for all 4-stair numbers on any size grid.

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 |

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 -------- ①

Term 2: a(2)3+ b(2)2 + c(2)+ d = 1

8a + 4b + 2c + d = 1 -------- ②

Term 3: a(3)3+ b(3)2 + c(3)+ d = 4

27a + 9b + 3c + d = 4 -------- ③

Term 4: a(4)3+ b(4)2 + c(4)+ d = 10

64a + 16b + 4c + d = 10 -------- ④

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.

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