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  • Level: GCSE
  • Subject: Maths
  • Word count: 4730

Number stairs.

Extracts from this document...

Introduction

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Introduction

Below you will find a 10 by 10, 9 by 9, 8 by 8, 7 by 7 and 6 by 6 number grids all of whom who will show 5 3 step stairs. For each 3 step stair I will workout the total of them and then will go on to find where appropriate algebraic formulas. Once I have done this I will go on to see whether or not the formula which I have worked out for a 10X10 number will work on another sized grid and one all the grids have been completed I will create a table of results looking at the formulas which I have found and on what sized grid they belong to. It will be best to display my results on a table as it will be easier for me to read off my formulas.

Part 1

For part 1 will investigate the relationship between the stair total of the 3 step stairs and the position of the stairs on the number grid.

10X10 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

Stair 1:

73 + 74 + 75 + 83 + 84 + 93 = 482

image209.pngimage142.png

image239.pngimage249.pngimage231.pngimage02.png

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 A+A+1+A+2+A+10+A+10+1+A+20=

 6A+44=482

 6A=482-44

 6A=438÷77=6

 A=73

 6(73) +44

Stair 2:

77 + 78 + 79 + 87 + 88 + 97 = 506

image11.pngimage10.png

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image16.pngimage15.png

image17.pngimage19.pngimage18.png

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X+X+1+X+2+X+10+X+10+1+X+20 =

6X+44=506

6X=506-44

6X=462÷77=6

X=77

6(77) +44

Stair 3:

21 + 22 + 23 + 31 + 32 + 41 = 170

image23.pngimage25.png

image34.pngimage26.png

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image72.pngimage77.pngimage86.png

Y+Y+1+Y+2+Y+10+Y+10+1+Y+20=

6Y+44=170

6Y=170-44

6Y=126÷21=6

Y=21

6(21) +44

Stair 4:

25 + 26 + 27 + 35 + 36 + 45 = 194

image96.pngimage92.png

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image133.pngimage35.pngimage47.png

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L+L+1+L+2+L+10+L+10+1+L+20=

6L+44=194

6L=194-44=150

6L=150÷25=6

L=25

6(25) +44

Conclusion:

Based on all of the 4 3 step stairs that I have investigated, I can now say that I have found a formula which works with all of the other 3 step stairs. The formula which I have found is 6A+44 which just simply means 6AX4X10+4.

...read more.

Middle

Grid Size:

Algebra:

General Formula:

10X10

         6Y+44 – stair 1

6Y X 4 X 10 + 4

9X9

6B+40 – stair 1

         6B X 4 X 9 + 4

8X8

6F+36 – stair 1

         6F X 4 X 8 + 4

7X7

6J+32 – stair 1

         6J X 4 X 7 + 4

6X6

6N+28 – stair 1

         6N X 4 X 6 + 4

From my table of results on the previous page, I have noticeda general trend:

I have noticed that as that grid size depreciates e.g. by one size from a 10X10 to a 9X9, the algebra also decreases by 4. This is because the grid size is getting smaller and therefore the numbers inside the grid are being reduced from 100 to 81.

 13

  14

 15

 16

  9

  10

  11

  12

  5

  6

  7

  8

  1

   2

3

  4

  • I will now test my theory that the size of the grid  will depend the outcome of the algebra and the algebra will be 4 less then its predecessor e.g. the algebra changing from 6n + 28 on a 6X6 grid to 6n + 20 on a 4X4 grid. I will prove this by drawing up a 4X4 grid. The ‘Letter’ value in the grid is ‘n’.

1 + 2 + 3 + 5 + 6 + 9 = 26

n+n+1+n+2+n+4+n+4+1+n+8=

6n+20=26

6n=26-20=6

6n=6÷1=6

n=1

6(1) + 20

My test worked. It shows that the size of the grid always shows the outcome of the algebra and most importantly I have noticed the trend which has reduced to 20 as I predicted.

Extending my Original line of Enquiry:

Now that I have finished my 3 step stairs on different grid sizes, I will now extend my line of enquiry by investigating the

...read more.

Conclusion

The formula for triangle numbers is:

nth term = ½ n (n+1), this is the same as my formula when ‘n’ and ‘x’ mean the same thing:

½ X (x+1)

Testing my formula:

½ x² + ½ x = ½ X (x+1)

when x = 4,  ½ + 16 + ½ X 4 =10

8 + 2 = 10

Pyramid Numbers

The 3rd differences are the same, so as a result it is cubic. The cubic formula is:

a X³ + b X ² + c X + D

Substitute:

1x = 1:                        A + B + C + D = 0

2x = 2:                       8A + 4B + 2D + D = 1

3x = 3:                       27A + 9B + 3C + D =4

4x = 4:                       64A + 16B + 4C + D = 10

8A + 4B + 2C + D = 1                      64A + 16B + 4C + D = 10

A + B + C + D = 0                             27A + 9B + 3C + D = 4

7A + 3B + c = 1                                37A + 7B + C = 6

eliminates D:  27A + 9B + 3C + D = 4

                         8A + 4B + 2C + D = 1

                         19A + 5B + C = 3

19A + 5B + C = 3                                     37A + 7B + C = 6

7A + 3B + C = 1           eliminates C      19A + 5B + C = 3

12A + 2B = 2                                            18A + 2B = 3

to get B:

18A + 2B = 3                           substitute: 12A + 2B = 2

12A + 2B = 2                           a = ¹/6      12x ¹/6 + B = 2

6A = 1                                      12x ¹/6 + 0 = 2

A = ¹/6                                     B = 0

to get C:                                    now we put our A, B + C values

                                                  into the cubic number formula:

Substitute: 7A + 3b + C = 1                               a =   ¹/6:     a X ³ + b + x² + CX + d

A = ¹/6        7x ¹/6 + 3x D + C = 1                     b = 0 :         ¹/6 x³ + - ¹/6X + 0

B = ¹/6        1¹/6 + D + C = 1                             c = - ¹/6:     ¹/6 X³ + - ¹/6 X

                                                                              d = 0

13

14

15

16

9

10

11

12

5

6

7

8

1

2

3

4

Now I am going to test my findings on my general rule, which is:

(½ ² + ½ ) n + (¹/6 X² = ¹/6x) g+ (¹/6 X² - ¹/6X)  +  (½ X 4² + ½ 4) + ¼ X 4 = 60

x = 4

n = 1

g = 4

My general rule for any grid size with any number of stairs has been proven to work.

Why you get triangle numbers and Why you get pyramid numbers:

image221.png

image222.pngimage223.png

image225.png

1 Step Stair = 1n, 2 Step Stair = 3n + g + 1,  3 Step Stair = 6n + 6g + 4, 4 Step Stair = 10n +10g + 10

image227.png

Candidate Name: Amish Patel Candidate Number: 8153   Centre Number:  12504

...read more.

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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