n: 1 2 3 4 5
50 56 62 68 74
Thus the nth term rule is 6n + 44. This is because as you move to each position the total of the numbers increase by six as there are six squares.
The next results should show again if there is a relationship between the stair total and the position of the stair shape vertically on the number grid. I shall also investigate this by recording 5 different positions on the number grid. I will begin with number 25 on the number grid up to number 65, by using the bottom left hand corner of the 3-step stair as an indication of which position the 3-step stair is in.
As you can see there is also a linear sequence between the total of the 3-step stairs and the position that they are in:
194 254 314 374 434
The relationship that is seen is that the rule for moving across from term to term is +60.
Formulae
To calculate a formula I need to call the bottom left corner n and the square next to it is n+1 and so on, as shown below:
To create a formula I will add the algebraic letters. I have used algebra as it would have taken a large amount of time to use numbers and algebra can give a formula that can work for every position possible in the number grid. Thus algebra can prove that the formula works;
Formula: (n) + (n+1) + (n+2) + (n+10) + (n+11) + (n+20)
= 6n + 44
Comments on Formulae
Basically the 6 in the formula comes form the number of squares in the 3-step stair and the 44 comes from this quadratic equation;
Position: 1 2 3 4 5
Total: 50 56 62 68 72
Total = an + b Therefore a= 6
6+b = 50
b = 50-6
b = 44
Therefore, 6n+44 can be used as a formula for every position possible on the number grid.
Testing my Predictions
To prove that my formula works, I can test my predictions by firstly manually calculating the total of the 3-step stair, and then algebraically calculating the total using my formula finally I will randomly select three positions to test. If my formula works I should have the same total for both of the stairs;
- Position 58 = 58 + 59 + 60 + 68 + 69 + 78 = 392
By using my formula (6n + 44), I should get the same total, thus if n = 58,
6 x 58 + 44 = 392
- Position 16 = 16 + 17 + 18 + 26 + 27+ 36 = 140
By using my formula (6n + 44), I should get the same total, thus if n = 16,
6 x 16 + 44 = 140
- Position 32 = 32+ 33 + 34 + + 27+ 36 = 140
By using my formula (6n + 44), I should get the same total, thus if n = 16,
6 x 16 + 44 = 140
Plan for Part 2
For part 2 of this task I shall investigate further the relationship between the stair totals and other step stairs on other number grids. To do this, I will change the size of the number stair (from a 1-step stair to a 5-step stair) and calculate the total of the algebraic letters inside it. This can indicate what the relationship between the totals of the number stairs are compared to their sizes. Then I will work out an overall formula for the first part of the formula (e.g. 6n) for other step stairs on a 10 by 10 grid and afterwards I will work out an overall formula for the second part of the formula (e.g. +44) for other step stairs on a 10 by 10 grid. I will finally combine the two formulas to create a final formula for working out the formula for any number stair. To conclude I shall further investigate other number stairs on different number grids.
Prediction
I predict that a formula can be derived for any number stair. Consequently, I will see a pattern in both parts of the formula that I obtained in part 1.
Results
I shall first investigate what the formulas are for other step stairs, this is because I want to find an overall formula for any step stair possible on a 10 by 10 grid and I can do this by comparing it to other step stairs;
1-Step stair
Total: n
Formula: 1n
2-Step stair
Total: n + n + 1 + n + 10
Formula: 3n + 11
3-Step stair
Total: n + n + 1+ n + 2 + n + 10 + n + 11
n + 20
Formula: 6n + 44
4-Step stair
Total: n + n + 1+ n + 2+ n + 3 + n + 10 + n + 11+ n + 12+ n + 20 + n + 21 + n + 30
Formula: 10n + 110
5-Step stair
Total: n + n + 1+ n + 2+ n + 3 + n + 4 + n + 10 + n + 11+ n + 12+ n + 13 + n + 20 + n + 21+ n + 22 + n + 30 + n + 31+ n + 40
Formula: 15n + 220
To make the results easier to read I have placed this into a table:
To create a formula for the total of the step-stairs I will use a quadratic sequence.
Stairs (s): 1 2 3 4 5
Total (t): a + b + c 4a + 2b + c 9a + 3b + c 16a+4b+c 25a+5b+c
1st difference 3a + b 5a+ b 7a +b 9a + b
2nd difference 2a 2a 2a
I can work out the value of a: 2a = 1
a= 1/2
a = 0.5
I can work out the value of b: 3a + b = 2
(3 x 0.5) + b = 2
1.5 + b = 2
b= 2-1.5
b = 0.5
I can work out the value of c by: a +b + c =1
0.5 + 0.5 + c = 1
c = 1- 1
c = 0
So the formula for the first part is t = (s2 + s)
2
To calculate the second part of the formula I have found a cubic sequence.
Number stairs: 1 2 3 4 5
Totals: 0 11 44 110 220
1st difference 11 33 66 110
2nd difference 22 33 44
3rd difference 10 10
The general formula for cubic ax3 + bx2 + cx + d, where a, b, c and d are constants and x is the variable. Thus I can insert my values into it to find the second part of the formula.
Step stairs: 1 2 3 4 5
Totals: a+b+c+d 8a+4b+2c+d 27a+9b+3c+d 64a+16b+4c+d 125a+25b+5c+d
1st difference 7a+3b+c 19a+5b+c 37a+7b+c 61a+9b+c
2nd difference 12a+2b 18a+2b 24a+2b
3rd difference 6a 6a
To find the value of a: 6a = 11
a= 11/6
a = 1.83
To find the value of b: 12a + 2b = 22
22 + 2b = 22
2b = 22-22
b =0/2
b = 0
To find the value of c: 7a + 3b+c= 11
12.83 + c = 11
c= 11-12.83
c = -1.8
To find the value of d: a + b + c + d = 0
1.8 – 1.8 + d = 0
d = 0
Consequently, I have now calculated a formula for the second part; 11s3 – 11s
6
As I have found the two formulas I can combine them together to form a single formula for the total of any number stair.
Total: (s2 + s) + 11s3 – 11s
2 6
Testing my predictions
To prove that my formula works I can test my predictions by seeing if my number stair formula is calculated to be the same as the formulas that I have already calculated before;
2-Step stair
Total: (s2 + s) + 11s3 – 11s
2 6
(22 + 2) + 11x (23) – 11x 2
2 6
6 + 88 – 22 = 3n + 11
2 6
3-Step stair
Total: (s2 + s) + 11s3 – 11s
2 6
(32 + 3) + 11x (33) – 11x 3
2 6
12 + 297 – 33 = 6n + 44
2 6
4 -Step stair
Total: (s2 + s) + 11s3 – 11s
2 6
(42 + 4) + 11x (43) – 11x 4
2 6
10 + 704 – 44 = 10n + 110
2 6
5-Step stair
Total: (s2 + s) + 11s3 – 11s
2 6
(52 + 5) + 11x (53) – 11x 5
2 6
30 + 1375 – 55 = 15n + 220
2 6
Conclusion
The aim of this coursework was to first investigate for other 3-step stairs, the relationship between the stair total and the position of the stair shape on the grid. Then to investigate further the relationship between the stair totals and other step stairs on other number grids.
I have completed part 1 of the task by thoroughly investigating the relationship between the totals of the numbers inside the 3-step stairs and their position horizontally and vertically on the grid. Furthermore, by inserting algebraic numbers into a 3-step stair and adding the algebraic letters together, I have found that the formula 6n + 44 links the total of numbers inside any 3-step stair to any possible position on a number grid.
For the part 2 of the task, I have changed the size of the number stair, starting with a 1-step stair and stopping with to a 5-step stair. Then by calculating the total of the algebraic letters inside each step stair and using quadratic and cubic sequences I have been able to calculate an overall formula for any number stair. This was as follows:
Total: (s2 + s) + 11s3 – 11s
2 6
This then allowed me to find a pattern between the first (e.g. 6n) and second part (e.g. +44) of the overall formula written above.
I have subsequently proved the formula for part one i.e. 6n +44 by testing it for 5 possible random positions (vertically and horizontally) on the number grid. I have then repeated this for the second part of the task. This time, I proved that the overall formula could work by using the original formulas that I had first calculated for a 1-step stair up to a 5-step stair as a key. I then manually inserted the numbers required into the overall formula to get a result.
Extending the Investigation
There are a number of possible ways to extend this investigation. For example, it is possible to determine formulas for any step stair on say an 8 by 8 grid and then comparing them with those from a 10 by 10 grid. This can be further extended to finding an overall formula for any step stair on say an 8 by 8 number grid as shown below:
I shall first investigate what the formulas are for other step stairs in order to find an overall formula for any step stair possible on an 8 by 8 grid. I then compare this to other step stairs.
1-Step stair
Total: n
Formula: 1n
2-Step stair
Total: n + n + 1 + n + 9
Formula: 3n + 10
3-Step stair
Total: n + n + 1+ n + 2 + n + 9 + n + 10
n + 17
Formula: 6n + 39
4-Step stair
Total: n + n + 1+ n + 2+ n + 3 + n + 9 + n + 10 + n + 11+ n + 17 + n + 18 + n + 25
Formula: 10n + 96
5-Step stair
Total: n + n + 1+ n + 2+ n + 3 + n + 4 + n + 9 + n + 10+ n + 11+ n + 12 + n + 17 + n + 18+ n + 19 + n + 25 + n + 26 + n + 33
Formula: 15n + 190
By looking at the number stairs I can see another pattern similar to the numbers on a 10 by 10 grid.
10 by 10 grid 8 by 8 grid
