Staircase Coursework

6x10 + 28 = 88

So the formula is correct

                               3 step stair in a variable grid

I’ve found the formula for the total number of 3 step stair on a 10x10 grid, and the formula for a 3 step stair on a 6x6 grid,  but I can not use this formula to calculate a 3 step stairs on a variable grid.

So I will now try to investigate a formula, for a 3 step stair on a variable grid.

I will again use a 6x6 grid for my investigation to keep it simple.

A 3-step stair on 6x6 stair case

                 

n = stand for size of the grid ( bottom left number )

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

I will now replace the number in the square by using n, 1 and and if possible g.

For example 8 is : n + g + 1  because it basically means 1 + 6 + 1 which equals 8.

But in square 3 you can for example not insert g because it is so big. Therefore I can just use n and g which would give me for 3:      3 = n+2

You now have to use this principle for all squares.

My result is shown below

   

             

 

       =

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So I can now add up the different squares, which gives me the formula:

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

If I now simplify this result I get the end formula

                              St = 6n + 4g + 4

I am now going to test the formula to see if it is correct.

So I again just add up all the numbers in the squares of another example than stair 1. For example step10, which gives me:

10+11+12+16+17+22 = 88

 And I now just insert :  n = 1   and  g = 6 into the formula St = 6n + 4g +4

6 x 10 + 4 x 6 + 4 = 88

So the formula is correct

Now I am going to test the formula a second time on a fare bigger grid to check if it is still correct.

                                    3 step chair on a 15 x 15 grid

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

      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= 46     St = 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

                 

I will again use step 1 to find out the formula which gives me

n =1  and   g = 15     because the left bottom number is 1 and the grid size is 15

I now insert n and g for the each number in the squares, which gives me the long formula:

   

I will now again add up all the g`s, n`s and numbers which gives me:

This simplified formula:

St= 10n+10g+10

I will now test if the formula by looking at stair 11.

By adding up all the different squares I get.

11+12+13+14+26+27+28+41+42+56 = 270

And If I now use the formula and insert n = 11  and   g = 15  

I get:

10x11 + 10x15 + 10 = 270

So the formula is correct

                              2 step stair on a variable grid

              I will now try to find a formula for a 2 step stair on a variable grid.

A 2-step stair on 6x6 stair case

                 

So again because n = 1   and   g = 2 for stair 1 I get the formula:

n+(n+1)+(n+gx3)

So this formula simplified equals:

St = 3n + 3g + 1

I Iii#

II will now test the formula to see if it works, by looking at stair 16

So if  n = 16   and g = 2

3x16 + 3x2 + 1 = 55

 

And if I add up the single squares I get

16 + 17 + 22 = 55

So the formula works

Perfect overall formula

What I am going to try now, is to find a formula for a variable size of grid and a variable size of stairs

First I will find out the relationship between the numbers.

If stair number = n

Grid size = g

I will list the sum of 2 steps stairs, 3 steps stairs and 4 step stairs in an equation from.

2x2...   3n + 3g + 1

3x3…  6n + 4g + 4

4x4… 10n+ 10g+10

I see the structure of these formulas; I will split the formula into 3 parts

Part 1 is the number of square in the staircase multiplies by the stair number.

Part 2 is the number of “n”s in the staircase multiply by the grid site.

Part 3 is the number of “1”s in the staircase.

I will try to work out a formula for Part 1 first…

Part 1 is the total number of squares in the staircase

For example for the case of a 4 step stair, there are 10 squares:

 

You should notice, that the staircase is a triangle!!!

So logically I decided to use the triangle number formula to calculate the total of squares in the staircase.

 

How did I find out this formula for a Triangular number:

The image above pictures each integer as a row of dots.

The first triangular number is 1

The second triangular number is 1 + 2 = 3

The third triangular number is 1 + 2 + 3 = 6

The forth triangular number is 1 + 2 + 3 + 4 = 10

So the x th triangular number is 1 + 2 +3 + ........+ x

Therefore the first triangular numbers are 1, 3, 6, 10

To get the formula for the x th triangular number, the sum must be found writing the sequence twice as follows.

St = 1 + 2 + 3 + 4 + ...... + x

Writing this reverse

St = x + ...... + 4 + 3 + 2 + 1

Now I add these 2 sums up, which gives me

2xSt = (x+1)+(x+1)+.....(x+1)+(x+1)             n times

2xSt = x (x+1)

St  =  1/2 x (x+1)              

1/2x (x+1)        

So the formula of triangle number: 1/2x (x+1)              x= size of triangle/stair (e.g. 6 step stair x = 6 )

I now have to multiply “n” (stair number) to the formula because according to the relation of the number in the staircase, every square contains a stair number, which would be as follows.

So I get the first part of the formula!!!

[1/2x (x+1)] n                 n = left bottom number  

                                       x = number of steps

 I will now work out the second and the third parts.

I notice that 1, 4, 10, ……….. are tetrahedral numbers, so I will use the formula of the tetrahedral numbers, which is: 1/6x (x+1) (x+2)

I got this formula out of the Internet

Tetrahedral numbers or triangle pyramidal numbers are figurate numbers that represent a pyramid with a triangular base and free sides called tetrahedron.

The tetrahedral numbers are:

1, 4, 10, 20, 35, 56, 84, 120 ........

The sequence of the stairs on my table and the sequence of the tetrahedral numbers, are not the same, the sequence of the stairs is “1” more then the sequence of tetrahedral numbers, so I will subtract the unknown in the formula by 1

1/6(x-1) (x-1+1) (x-1+2)           1/6(x-1) (x) (x+1)

Now I have to multiply the whole formula by g, which is the grid size because part 2 stands for the total of n `s multiplied by the grid size

So the overall second part of the formula is [1/6(x-1) (x) (x+1)] g

Finally, the third part

I listed part 2 and part 3 on the table.

I notice that the number of Part 2 is equal to the number of Part 3

So I can use the formula of part 2 again: 1/6(x-1) (x) (x+1) 

But I will obviously not multiply this part of the formula by g, because part 3 just shows the 1 `s and because there are as many 1´s as n ´s I can use the formula again.

        

Now, I will add part 1, 2 and 3 together.

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

I will now use word equation to explain the formula to make it easier to understand.

[Total of square in stair case] stair number + [number of “n”s in the stair case]grid size + [number of “1”s in the stair case]

I will now simplify the formula in order to make it easier to calculate

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

[1/2x (x+1)] n + (g+1) [1/6(x3-x)]                              this is the formula!!!

I will test the formula to look if it’s correct:

                                          8 step stair on a 10x10 grid

Now I will use the formula to calculate the total…

So in this case :

x=8

g=10

n=13

Using formula:

[1/2x (x+1)] n + (g+1) [1/6(x3-x)]

  [1/2x8 (8+1)] 13 + (10+1) [1/6(83 - 8)]

=[4x9] 13 + 11 [84]

=36 x 13 +11 x 84

=468 + 924

=1392

I will now test the formula by adding up all the single squares, which gives me:

S13: 13+14+15+16+17+18+19+20+23+24+25+26+27+28+29+33+34+35+36+37+38+43+44+45+46+47+53+54+55+56+63+64+65+73+74+83 = 1392

So the formula is correct

So the end formal for a variable size grid and a variable size step is :

[1/2x (x+1)] n + (g+1) [1/6(x3-x)]

                                Conclusion:

At the beginning I had some difficulties, and made mistakes.

I used the same example to test the formula, like I already used to come up with it. This is obviously not correct, because if you use the same step you already used, it will always be correct even if you made a mistake.

I could imagine if I repeat the coursework it would take me far less time, but in the end the results would be the same, because I worked very accurate and precise.

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(x3-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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