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• Level: GCSE
• Subject: Maths
• Word count: 4685

# Maths coursework. For my extension piece I decided to investigate stairs that ascend along with the numbers, in order to do this the grid was turned upside-down. I aim to see if there is a pattern within these stairs

Extracts from this document...

Introduction

For my extension piece I decided to investigate stairs that ascend along with the numbers, in order to do this the grid was turned upside-down. I aim to see if there is a pattern within these ‘stairs’

 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

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

Middle

= 21 + 5 + 1

= 27

Stair-total (found by adding) = 7 + 8 + 12 = 27

This shows that my formula must work for all 2-stair numbers on the 5 x 5 grid.

However I was yet to find out if this formula would also work on other grid sizes. To investigate this, I am going to see if the formula still works for 2-stair shapes on an 8 x 8 grid.

 57 58 59 60 61 62 63 64 49 50 51 52 53 54 55 56 41 42 43 44 45 46 47 48 33 34 35 36 37 38 39 40 25 26 27 28 29 30 31 32 17 18 19 20 21 22 23 24 9 10 11 12 13 14 15 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

This must mean that my formula Un = 3n + g + 1 – where ‘n’ is the stair number, ‘g’ is the grid size, and ‘Unis 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

Conclusion

I have noticed that all the formulae are made up of three terms: an ‘n’ term, a ‘g’ term and a ‘number’ term (the first formula could be written as 1n + 0g + 0). These terms are always added together to form the formula. The coefficient of ‘g’ is always the same as the number in the ‘number’ term.

Therefore I can tackle this problem of finding a rule connecting the number of stairs and its formula for working out the stair total in two parts: finding the rule connecting the number of stairs and the coefficient of the ‘n’ term, and then finding a rule connecting the number of stairs and the coefficient of the second (and in effect the third term as well).

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                       --------

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

8a + 4b + 2c + d = 1                  --------

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

27a + 9b + 3c + d = 4                --------

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

64a + 16b + 4c + d = 10                        --------

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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