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Introduction     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            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            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            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            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.

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

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:    1 Step Stair = 1n, 2 Step Stair = 3n + g + 1,  3 Step Stair = 6n + 6g + 4, 4 Step Stair = 10n +10g + 10 Candidate Name: Amish Patel Candidate Number: 8153   Centre Number:  12504

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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