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  • Level: GCSE
  • Subject: Maths
  • Word count: 3422

Staircase Coursework

Extracts from this document...

Introduction

Dominik Borgolte

Math GCSE Coursework

Dominik Borgolte

Buckswood School

GCSE Mathematic coursework

“Step stairs”

Centre number:

Candidate number: 2776


Introduction

I am going to investigate formulas that can be used to calculate the sum of the number in a staircase on a grid paper.

In my investigation I will use:

St: stand for stair total ( S stand for stair and t for the total sum )

g: stand for the size of the grid (e.g. in 12x12 grid n=12)

n: stand for stair number, which is always the bottom left number

x. stand for the size of the stairs ( e.g. a 6 step stair would be x=6 )


3 step stair on a 10x10 grid

( 1 )        

I want to find a formula, which makes it easier for me to calculate the total size of a 3 step stair in a 10 x 10 grid.

10 x 10 grid

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100

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

I have added all the squares of the 3 step stairs together, which gives me a result of 50, for step1.

I will now find a formula for the total, which will make it easier to calculate the sum.

I will label a stair algebraically

This is an example of stair 1

n stands for the number in the left bottom corner of the step.

In order of that if : 1 = n          2 = n + 1     and so on ................

n+20

n+10

n+11

n

n+1

n+2

21

11

12

1

2

3

          =

...read more.

Middle

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Formula:  6n + 4g +4

g = 15

n = 97

6x97 + 4x15 + 4 = 646

Test of the formula:

St:  97+98+99+112+113+127=646

So the formula also works for any other size grid

( 3a )

4 step stair on a 10x10 grid

I will now try to find a formula for a 4 step stair on a 10 x 10 grid

                                              4 step stair on a 10x10 grid

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10

      I use the same principle like I already did with the 3 step stair on a 10x10 grid.

So I make the bottom left number n, which gives me a huge formula.

In this case

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

Then I add up all normal numbers and all the n`s

So in this case it would be 10 times n and all the numbers added up would be 110.

This leaves me with the final formula

St = 10n + 110

I will now test the formula to see if it is correct by using step 46

     Because n= 46St = 10x46 + 110= 570

                             St = 46+47+48+49+56+57+58+66+67+76 = 570

                            So the formula is correct

(3b)                4 step stair in a variable grid

Now I will try to find a formula for a 4 step stair on any grid.

I will use a 15x15 grid for my investigation

                4-step stair on 15x15 grid

46

47    

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...read more.

Conclusion

But still a few improvements could be made. I could for example increase the amount of tests on different sizes of stairs and grids

 I would try out a 5 or 8 step stair, or test them at different size grids like I already used, but in the end the final result, which is the overall formula  

[1/2x (x+1)] n + (g+1) [1/6(x[4]3-x)] would still be the same.

I had some help from my teacher and an older classmate. I also took one formula for the tetrahedral number out of the Internet. Other than that I did everything on my own, which is highly important in my opinion.

Therefore I think overall the coursework was a clear success, because I was able to find formulas for every different size step, and every different size grid investigation I made.

In the end I was even able to find a formula that can be used for any size grid and any size stair, so I could even achieve my final goal of the coursework.

I also think that this coursework was useful for my personal abilities, because I really improved my ability to think logically, and my ability to help my self when I am sucked at a certain point, which was the case (a few times) during this coursework.  


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...read more.

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