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  • Level: GCSE
  • Subject: Maths
  • Word count: 3157

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Introduction

Ahmed Alaskary

GSCE COURSEWORK                                  

PART 1

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I drew 5 3-step stairs all of them were the same size, but they differed in position.

Number Of Shape

Calculations

Total

1

1+2+3+11+12+21

50

8

8+9+10+18+19+28

92

34

34+35+36+44+45+54

248

71

71+72+73+81+82+91

470

78

78+79+80+88+89+98

512

Comment:

1) Notice how all the totals are even

2) The totals increase the higher up the grid you go

3) Across goes up about 42 (92-50)        

4) Up goes up about 420 (470-50)

Systematic Sample

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I drew 5 stairs all on the bottom line going across, I did this to see if there was a pattern.

Number Of Shape

Calculations

Total

1

1+2+3+11+12+21

50

2

2+3+4+12+13+22

56

3

3+4+5+13+14+23

62

4

4+5+6+14+15+24

68

5

5+6+7+15+16+25

74

Comment:

There is clearly a pattern here; the totals are increasing by 6, we call this a linear sequence.

From this information I predict that the next total will be 80 and the 8th will be 92.

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Results:

Number Of Shape

Calculations

Total

6

6+7+8+16+17+26

80

8

8+9+10+18+19+28

92

Both my predictions were correct.

Algebraic Expression

X+20

X+10

X+11

X

X+1

X+2

I drew 1 stair with Algebra on it.

Total:

(x)+(x+1) +(x+2) +(x+10) +(x+11) +(x+20) = 6X+44

Comment:

The reason it’s 6x is because there are 6 stairs (x+x+x+x+x+x)

It’s 44 because the totals across for these stairs are 4 acrossand 40 upwards.

4: 1+1+2=4

1: This 1 came from the 11, because you go 1 across from the 10 to get 11

40: 20+10+10=40

10: This 10 came from the 11, because you go 10 up from the 1 to get 11

PART 2

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...read more.

Middle

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Results:

Name Of Grid

Calculations

Total

9

1+2+3+10+11+19

46

My prediction was correct.

Algebraic Expression

x+g+g

x+g

x+g+1

x

x+1

x+1+1

I have drawn a 3-step stair with algebra on it, the g represents grid.

Totals:

(x)+(x+1) +(x+1+1) +(x+g) +(x+g+1) +(x+g+g) = 6x+4g+4

Comment:

The reason it’s 6x is because there are 6 stairs, each one with one x on it (x+x+x+x+x+x=6x), x represents the name of the shape.

The reason it’s 4g is because say for example you had a 5 grid, notice how every time you go up the total increases in accordance with the grid number; e.g.:

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image00.png

image00.pngimage00.png

                 +5

So basically if it were any grid the number on top of one would always increase by the grid number. That is why the g is situated on top of the x in my algebra stairs; and since there are 4g’s we write it down in the formula as 4g.

The reason it’s +4 is simply because it’s (1+1+1+1). These ‘1’s’ are on my algebra stairs because in any stairs no matter what grid it is, you always go one across.

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image01.png

        +1

So, my formula is: 6x+4g+4

I will now experiment it on 3 different grids to see if it works.

I will try it on a 3 grid, a 7 grid and a 10 grid.

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Name Of Grid

Calculations

Total

3

1+2+3+4+5+7

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Name Of Grid

Key

Calculations

Total

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x=1        g=3

(6x1)+(4x3)+4

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Name Of Grid

Calculations

Total

7

5+6+7+12+13+19

62

Name Of Grid

Key

Calculations

Total

7

x=5       g=7

(6x5)+(4x7)+4

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Name Of Grid

Calculations

Total

10

45+46+47+55+56+65

314

Name Of Grid

Key

Calculations

Total

10

x=45       g=10

(6x45)+(4x10)+4

314

My formula worked for all 3 grids, from this I can conclude that the formula, 6x+4g+4 will work on any grid provided that it’s a 3 step-stair.  

The formula 6x+4g+4 is a linear formula. I have found a formula that will work on any 3-step stair, now I need to find a formula which will work on any sized step.

I will now change the stairs size to try and find a formula which will work on any stair size.

I will do a 3, 4, 5 and 6 step stairs on a 10 grid.

                           3-STEP STAIR

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...read more.

Conclusion

6

10 x 3= 3 x 10          Notice how these are triangle numbers. However, it’s the value of n+1.

20 x 3= 4 x 15          So we substitute it with the triangle numbers formula except we use (n+1):

35 x 3= 5 x 21          [n (n+1) (n+1) +1)]/2, but remember we have to divide by 3 because we

                                 multiplied by 3 earlier so the formula is:  [n (n+1) (n+2)]/6

                                 In our series the 4 is from a 3 step, the nth step would need to be (n-1) in          

                                 the above formula (Blue). We substitute n with (n-1):

                                                 [(n-1) ((n-1) +1) ((n-1) +2)]/6

                                 We can simplify the formula into this:

                                                 [n (n+1) (n-1)]/6

So the universal formula is:   [n (n+1)]/2 x + [n (n+1) (n-1)]/6 g+ [n (n+1) (n-1)]/6

                                               n= step size

x= number of shape

                                               g= grid number

I will now test to see if it works, I want to get the formula for a 3-step stair, so I do the following:

[3(3+1)]/2 x + [3(3+1)(3-1)]/6 g + [3(3+1) (3-1)]/6  which equals:

6x+4g+4 – My formula works for the 3-step stairs.

I will do the same, but this time for a 7 step stairs:

[7(7+1)]/2 x + [7(7+1) (7-1)]/6 g + [7(7+1) (7-1)]/6 which equals:

28x+56g+56

Clearly my formula works, through the use of algebra and my insight into the relationships of series and triangle numbers I was able to achieve this. I needed to use a variety of methods and techniques to complete the universal formula, always linking it back with previous formulas and tables.

...read more.

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