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  • Level: GCSE
  • Subject: Maths
  • Word count: 4455

Mathematics - Number Stairs

Extracts from this document...

Introduction

Ben Foster        GCSE Mathematics Coursework 02/02/2008

Ben Foster

-------- School---------

Centre Number --------

GCSE Coursework: Number Stairs

Teacher: -------


Table of Contents

Aim

First Part

3 Step-Staircase / Grids of width 10

Second Part

3 Step-Staircase

3 Step-Staircase / Grid Width 8

3 Step-Staircase / Grid Width 9

3 Step-Staircase / Grid Width 11

3 Step-Staircase / Grid Width 12

2 Step-Staircase

2-Step Staircase/ Grid Width 8

2-Step Staircase/ Grid Width 9

2-Step Staircase/ Grid Width 10

2-Step Staircase/ Grid Width 11

2-Step Staircase/ Grid Width 12

1 Step-Staircase

4 Step-Staircase

4 Step-Staircase / Grid Width 8

4 Step-Staircase / Grid Width 9

4 Step-Staircase / Grid Width 10

4 Step-Staircase / Grid Width 11

4 Step-Staircase / Grid Width 12

5 Step-Staircase

5 Step-Staircase / Grid Width 8

5 Step-Staircase / Grid Width 9

5 Step-Staircase / Grid Width 10

5 Step-Staircase / Grid Width 11

5 Step-Staircase / Grid Width 12

The Final Formula

Finding the “n” term of the formula

Finding the “g” term of the formula

Finding the number term of the formula

Aim

In this investigation, I will be aiming to find out the formulas of the combinations of the size of number stairs and the size of grid widths. Eventually I will convene a formula connecting the stair totals and other steps on other number grids.

First Part

3 Step-Staircase / Grids of width 10

image00.png

21

11

12

1

2

3

n

1

2

3

4

5

T

50

56

62

68

74

There is a pattern here, so a formula may be constructed:

It is going up by 6 every time thus making it “6n”

If the 1st term is 50 then the 0th term is 44

So therefore the formula here is T = 6n+44

To make sure this formula works, the n term can be substituted with a number, for example, 20

T = 6 x 31 + 44 = 230

So:

51

41

42

31

32

33

31 + 32 + 33 + 42 + 41 + 51 = 230

So therefore my prediction works but I must prove it algebraically:

n+20

n+10

n+11

n

n+1

n+2


n + (n+1) + (n+2)

...read more.

Middle

12

1

 T = n

 T = n

 T = n

T = n 

T = n 

2

 T = 3n + 9

T = 3n + 10 

 T = 3n + 11

 T = 3n + 12

T = 3n + 13

3

T = 6n + 36 

 T = 6n + 40

T = 6n + 44 

T = 6n + 48 

T = 6n + 52 

4

5

4 Step-Staircase        

4 Step-Staircase / Grid Width 8

25

17

18

9

10

11

1

2

3

4

n

1

2

3

4

5

T

100

110

120

130

140

Suspected formula: T = 10n + 90

Prediction / Test: 10 x 25 + 90 = 340

49

41

42

33

34

35

25

26

27

28

25 + 26 + 27 + 28 + 33 + 34 + 35 + 41 + 42 + 49 = 340

Algebraic Proof:

n+24

n+16

n+17

n+8

n+9

n+10

n

n+1

n+2

n+3

n + (n+1) + (n+2) + (n+3) + (n+8) + (n+9) + (n+10) + (n+16) + (n+17) + (n+24) = 10n + 90


8

9

10

11

12

1

 T = n

 T = n

 T = n

T = n 

T = n 

2

 T = 3n + 9

T = 3n + 10 

 T = 3n + 11

 T = 3n + 12

T = 3n + 13

3

T = 6n + 36 

 T = 6n + 40

T = 6n + 44 

T = 6n + 48 

T = 6n + 52 

4

 T = 10n + 90

5

4 Step-Staircase / Grid Width 9

28

19

20

10

11

12

1

2

3

4

n

1

2

3

4

5

T

110

120

130

140

150

Suspected formula: T = 10n + 100

Prediction / Test: 10x 24 + 100 = 340

51

42

43

33

34

35

24

25

26

27

24 + 25 + 26 + 27 + 33 + 34 + 35 + 42 + 43 + 51 = 340

Algebraic Proof:

n+27

n+18

n+19

n+9

n+10

n+11

n

n+1

n+2

n+3

n + (n+1) + (n+2) + (n+3) + (n+9) + (n+10) + (n+11) + (n+18) + (n+19) + (n+27) = 10n + 100

8

9

10

11

12

1

 T = n

 T = n

 T = n

T = n 

T = n 

2

 T = 3n + 9

T = 3n + 10 

 T = 3n + 11

 T = 3n + 12

T = 3n + 13

3

T = 6n + 36 

 T = 6n + 40

T = 6n + 44 

T = 6n + 48 

T = 6n + 52 

4

 T = 10n + 90

 T = 10n + 100

5


4 Step-Staircase / Grid Width 10

31

21

22

11

12

13

1

2

3

4

n

1

2

3

4

5

T

120

130

140

150

160

Suspected formula: T = 10n + 110

Prediction / Test: 10 x 23 + 110 = 340

53

43

44

33

34

35

23

24

25

26

23 + 24 + 25 + 26 + 33 + 34 + 35 + 43 + 44 + 53 = 340

Algebraic Proof:

n+30

n+20

n+21

n+10

n+11

n+12

n

n+1

n+2

n+3

n + (n+1) + (n+2) + (n+3) + (n+10) + (n+11) + (n+12) + (n+20) + (n+21) + (n+30) = 10n + 110

8

9

10

11

12

1

 T = n

 T = n

 T = n

T = n 

T = n 

2

 T = 3n + 9

T = 3n + 10 

 T = 3n + 11

 T = 3n + 12

T = 3n + 13

3

T = 6n + 36 

 T = 6n + 40

T = 6n + 44 

T = 6n + 48 

T = 6n + 52 

4

 T = 10n + 90

 T = 10n + 100

T = 10n + 110 

5

From here, I see a pattern of a jump size of 10 so I predict for grid widths 11 and 12 will be T = 10n + 120 and T = 10n + 130.


4 Step-Staircase / Grid Width 11

34

23

24

12

13

14

1

2

3

4

n

1

2

3

4

5

T

130

140

150

160

170

Suspected formula: T = 10n + 120

Prediction / Test: 10 x 47 + 120 = 590

80

69

70

58

59

60

47

48

49

50

47 + 48 + 49 + 50 + 58 + 59 + 60 + 69 + 70 + 80 = 590

Algebraic Proof:

n+33

n+22

n+23

n+11

n+12

n+13

n

n+1

n+2

n+3

n + (n+1) + (n+2) + (n+3) + (n+11) + (n+12) + (n+13) + (n+22) + (n+23) + (n+33) = 10n + 120

8

9

10

11

12

1

 T = n

 T = n

 T = n

T = n 

T = n 

2

 T = 3n + 9

T = 3n + 10 

 T = 3n + 11

 T = 3n + 12

T = 3n + 13

3

T = 6n + 36 

 T = 6n + 40

T = 6n + 44 

T = 6n + 48 

T = 6n + 52 

4

 T = 10n + 90

 T = 10n + 100

T = 10n + 110 

T = 10n + 120 

5


4 Step-Staircase / Grid Width 12

37

25

26

13

14

15

1

2

3

4

n

1

2

3

4

5

T

140

150

160

170

180

Suspected formula: T = 10n + 130

Prediction / Test: 10 x 15 + 130 = 280

51

39

40

27

28

29

15

16

17

18

15 + 16 + 17 + 18 + 27 + 28 + 29 + 39 + 40 + 51 = 280

Algebraic Proof:

n+36

n+24

n+25

n+12

n+13

n+14

n

n+1

n+2

n+3

n + (n+1) + (n+2) + (n+3) + (n+12) + (n+13) + (n+14) + (n+24) + (n+25) + (n+36) = 10n + 130

8

9

10

11

12

1

 T = n

 T = n

 T = n

T = n 

T = n 

2

 T = 3n + 9

T = 3n + 10 

 T = 3n + 11

 T = 3n + 12

T = 3n + 13

3

T = 6n + 36 

 T = 6n + 40

T = 6n + 44 

T = 6n + 48 

T = 6n + 52 

4

 T = 10n + 90

 T = 10n + 100

T = 10n + 110 

T = 10n + 120 

 T = 10n + 130

5


5 Step-Staircase        

5 Step-Staircase / Grid Width 8

33

25

26

17

18

19

9

10

11

12

1

2

3

4

5

n

1

2

3

4

5

T

195

210

225

240

255

Suspected formula: T = 15n + 180

Prediction / Test: 15 x 34 + 180 = 690

66

58

59

50

51

52

42

43

44

45

34

35

36

37

38

34 + 35 + 36 + 37 + 38 + 42 + 43 + 44 + 45 + 50 + 51 + 52 + 58 + 59 + 66 = 690

Algebraic Proof:

n+32

n+24

n+25

n+16

n+17

n+18

n+8

n+9

n+10

n+11

n

n+1

n+2

n+3

n+4

n + (n+1) + (n+2) + (n+3) + (n+4) + (n+8) + (n+9) + (n+10) + (n+11) + (n+16) + (n+17) + (n+18) + (n+24) + (n+25) + (n+32) = 15n + 180

8

9

10

11

12

1

 T = n

 T = n

 T = n

T = n 

T = n 

2

 T = 3n + 9

T = 3n + 10 

 T = 3n + 11

 T = 3n + 12

T = 3n + 13

3

T = 6n + 36 

 T = 6n + 40

T = 6n + 44 

T = 6n + 48 

T = 6n + 52 

4

 T = 10n + 90

 T = 10n + 100

T = 10n + 110 

T = 10n + 120 

 T = 10n + 130

5

T = 15n + 180 


5 Step-Staircase / Grid Width 9

37

28

29

19

20

21

10

11

12

13

1

2

3

4

5

n

1

2

3

4

5

T

215

230

245

260

275

Suspected formula: T = 15n + 200

Prediction / Test: 15 x 39 + 200 = 785

75

66

67

57

58

59

48

49

50

51

39

40

41

42

43

39 + 40 + 41 + 42 + 43 + 48 + 49 + 50 + 51 + 57 + 58 + 59 + 66 + 67 + 75 = 785

Algebraic Proof:

n+36

n+27

n+28

n+18

n+19

n+20

n+9

n+10

n+11

n+12

n

n+1

n+2

n+3

n+4

...read more.

Conclusion

Staircase Size

1

2

3

4

Formula

0g+0

g+1

4g+4

10g+10

Row 1

               g+1          3g+3           6g+6

Row 2

                         2                   3

Row 3

1

After finding out the method of differences, it turns out to be exactly the same sequence for finding out the “g” term considering the differences between the staircase sizes of the number terms follow the same sequence as the “g” term!

So now, there is a constant difference of 1, therefore, 1 ÷ 6 = 1/6s³

Now, I would expect a quadratic but since the sequence is exactly the same as the “g” term formula, I am predicting an automatic linear sequence:

Staircase Size

1

2

3

4

1/6³ Formula

0-(1/6 x 1³)

1-(1/6 x 2³)

4-(1/6 x 3³)

10-(1/6 x 4³)

Formula Solution

[-1/6]

[-1/3]

[-1/2]

[-2/3]

Row 1

                    [-1/6]              [-1/6]               [-1/6]

As predicted, the second part to the formula is -1/6.

The overall product is now: TS = n(0.5s² + 0.5s) + g(1/6s³ - 1/6s) + (1/6s³ - 1/6s)image03.jpg

To neaten the formula up a bit:

This seems ludicrous at first so I will now substitute the letters with numbers to prove this formula works.

With “n” being 1, “s” being 3 and “g” being 10, I should come up with a total step number of 50.

image04.jpg

image05.jpgimage06.jpgimage07.jpgimage09.jpgimage08.jpg

Staircase Size = 3, Grid width = 10, bottom left number = 1

1 + 2 + 3 + 11 + 12 + 21 = 50

So I have seen that this works fully and now that I have the overall formula connecting the staircase sizes, the grid widths and the bottom left number of the staircases together; the total of the numbers in any staircase of any size of any grid width can be worked out by using this formula:image03.jpg

...read more.

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