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• Level: GCSE
• Subject: Maths
• Word count: 4076

# Step-stair Investigation.

Extracts from this document...

Introduction

William Murray

Step-stair Investigation

Cousework Submission 8th December 2003

For my GCSE Maths coursework I was asked to investigate the relationship between the stair total and the position of the stair shape on the grid. Secondly I was asked to investigate the relationship further between the stair totals and the other step stairs on other number grids. The number grid below has two examples of 3-step stairs. I will use Algebra as a way to find the relationship between the stair total and the position of the stair on the grid. I will use arithmetic and algebra to investigate the relationships between the grid and the stair further. The variables used will be:

Position of stair on grid = X

Sum of all the numbers within the stair = S

Step Size= n

Grid size= g

 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

The first thing I will do is find the formula for all 3-step stairs on a size 10 grid.

I started off by making the bottom left hand number X. X is also the position of the stair on the grid. So in the diagram coloured red above X=15. I then added up the rest of the numbers in the three-step stair in terms of X. So 16= X+1, 17=X+2, 25=X+10 etc. The 3 step-stair in terms of X looks like this:

 X+20 X+10 X+11 X X+1 X+2

If you simplify all the Xs and all the numbers you end up with this: 6X + 44, X+X+1+X+2+X+10+X+11+X+20=6X + 44. By investigating the formula above you will find that it is the formula for all 3-step stairs on a size 10 grid. I worked this out by adding together all the numbers in the 3-step stair and then using the formula to see if the formula comes up with the total of all the numbers in the 3-step stair.

Middle

11

12

13

14

1

2

3

4

5

6

7

By using the formula 10X+10g+10=S, I worked out the total of the numbers inside the blue area of the 4-step stair.

(10*18)+(10*7)+10= 180 + 70+ 10= 260

Then  added up the numbers in the 4-step stair as I did before.

18+19+20+21+25+26+27+32+33+39= 260

The two examples above prove that the formula 10x+10g+10 calculates the total of the numbers inside the area covered by a 4-step stair on any grid size.

5 step stairs:

 X+4g X+3g X+3g+1 X+2g X+2g+1 X+2g+2 X+g X+g+1 X+g+2 X+g+3 X X+1 X+2 X+3 X+4

By adding all the Xs, all the gs and all the numbers together I got:

X+X+1+X+2+X+3+X+4+X+g+X+g+1+X+g+2+X+g+3+X+2g+X+2g+1+X+2g+2+X+ 3g+X+3g+1+X+4g = 15X+20g+20. This is the formula for all 5-step stairs on any size grid.

To prove this formula works for all size grids and therefore works in general I drew two different sized grids and did the following calculations:

 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

By using the formula 15X+20g+20=S, I worked out the total of the numbers inside the green area of the 5-step stair.

(15*25)+(20*10)+20=595

Then I added all the numbers in the 5-step stair to see if it came up with the same answer:

25+26+27+28+29+35+36+37+38+45+46+47+55+56+65=595.

 21 22 23 24 25 16 17 18 19 20 11 12 13 14 15 6 7 8 9 10 1 2 3 4 5

By using the formula 15X+20g+20=S, I worked out the total of the numbers inside the turquoise area of the 5-step stair:

(15*1)+(20*5)+20=15+100+20=135

I then added up all the numbers in the 5-step stair to see if it gave the same number:

1+2+3+4+5+6+7+8+9+11+12+13+16+17+21=135

These two diagrams prove that 15X+20g+20=S on all sized grids.

6-step stairs:

 X+5g X+4g X+4g+1 X+3g X+3g+1 X+3g+2 X+2g X+2g+1 X+2g+2 X+2g+3 X+g X+g+1 X+g+2 X+g+3 X+g+4 X X+1 X+2 X+3 X+4 X+5

By adding all the Xs all the gs and all the numbers up together I got this:

X+X+1+X+2+X+3+X+4+X+5+X+g+X+g+1+X+g+2+X+g+3+X+g+4+X+2g+X+2g+1+X+2g+2+X+2g+3+X+2g+3+X+3g+X+3g+1+X+3g+2+X+4g+X+4g+1+X+5g = 21X+35g+35.

To prove that this formula works I drew up two grids and used the formula to calculate the total of the numbers inside the 6-step stair and saw if it was the right answer on both grids.

 57 58 59 60 61 62 63 64 49 50 51 52 53 54 55 56 41 42 43 44 45 46 47 48 33 34 35 36 37 38 39 40 25 26 27 28 29 30 31 32 17 18 19 20 21 22 23 24 9 10 11 12 13 14 15 16 1 2 3 4 5 6 7 8

I did this calculation to see if the formula works.

21X+35g+35 = (21*11) + (35*8) + 35 = 546

11+12+13+14+15+16+19+20+21+22+23+27+28+29+30+35+36+37+43+44+51= 546

 31 32 33 34 35 36 25 26 27 28 29 30 19 20 21 22 23 24 13 14 15 16 17 18 7 8 9 10 11 12 1 2 3 4 5 6

Conclusion

(3-1) = 2

Σ Tr =  T1 + T2 = 1+3 = 4

r = 1

So, for a 3 step stair the value of (blank)g + (blank) = 4. If you look up the formula for a 3-step stair is 6X+4g+4. I then proposed that the formula:

(n-1)           (n-1)

n(n+1)   +Σ   Tr =     g +     Σ  Tr =

2           r = 1                         r = 1

Is the formula for any step stair on any sized grid.

I tried it on the 4 step stairs I had investigated before. The 3-step stair, the 4-step stair, the 5-step stair and the 6-step stair.

So, by replacing n with 3, for a 3-step stair I get this:

(3-1)                                          (3-1)

3(3+1)  +  Σ   Τr = T1 + T2   g   +     Σ    Tr = T1+T2

2           r =1                                    r = 1

So by using this formula that I have explained above:

The formula gives: 6X+4g+4.  T1 + T2 = 4, (1+3).

This is the formula for a 3-step stair on any size grid. The formula works.

4-step stair:

(4-1)                                            (4-1)

4(4+1)  +  Σ   Τr = T1 + T2  + T3   g   +     Σ    Tr = T1+T2 + T3

2          r = 1                                             r = 1

So by using this formula that I have exlained above:

The formula gives: 10X+10g+10.  T1 + T2 + T3 = 10 (1+3+6)

This is the formula for a 4-step stair on any size grid. The formula works.

5-step stair:

(5-1)                                                 (5-1)

5(5+1)  +   Σ   Τr = T1 + T2  + T3 + T4  g   +     Σ    Tr = T1+T2 + T3 + T4

2            r = 1                                                   r = 1

By using the formula I have explained above;

The formula gives: 15X+20g+20. T1 + T2 + T3 + T4 = 20, (1+3+6+10)

This is the formula for a 5-step stair on any sized grid. The formula works.

6-step stair:

(6-1)                                                           (6-1)

6(6+1)  +   Σ   Τr = T1 + T2  + T3 + T4 + T5    g   +     Σ    Tr = T1+T2 + T3 + T4 + T5

2            r = 1                                                            r = 1

By using the formula I have explained above:

The formula gives: 21X+35g+35. This is the formula for a 6-step stair on any sized grid. The formula works.

I have now proved that the formula I found works for any step stair size on any size grid.

(n-1)          (n-1)

n(n+1)   +Σ    Tr =     g +      Σ     Tr =

2           r = 1                        r = 1

This concludes my first coursework submission. Submitted on:

8th December 2003.

William Murray

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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