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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’

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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

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The stair-total for this stair shape is 67 + 68 + 69 + 77 + 78 + 87 = 446

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The stair-total for this stair shape is 68 + 69 + 70 + 78 + 79 + 88 = 452

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The stair-total for this stair shape is 3 + 4 + 5 + 13 + 14 + 23 = 62

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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

...read more.

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.

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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 ‘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

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16

n + 2g

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20

11

n + g

n + g + 1

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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

...read more.

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                        --------

...read more.

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