- N=33 N=42
- 6 x 33 6 x 42
- 198+44 252+44
-
242 296
- 6n+44 6n+44
- N=51 N=41
- 6 x 51 6 x 41
- 306+44 246+44
-
350 290
- 6n+44
- N=32
- 6 x 32
- 192+44
-
236
To see if my formula is right I would have to test out my theory further using random 3-step stair shapes from the 10cm by 10cm gird. I am repeating the theory to ensure it’s accurate, precise and reliable for further investigation such as Part two.
I’ve shown that there does seem to be a relationship between the stairs total and the position of the stair shape. However to certify that such a link exist, I will continue my investigating by using the same formula. Nonetheless, I will use different shaped grids i.e. 6cm by 6cm to demonstrate that my formula is utilize in any circumstance. In addition, I will change the 3 step stairs into different steps such as; 4-step stairs. These changes enable me to investigate further and fully grasp the perception of the link between the stair shape and stair total. Nevertheless it will facilitate me to find the stair total quicker.
9cm by 9cm grid
The Calculations
1+2+3+10+11+19= 46
2+3+4+11+12+20= 52
3+4+5+12+13+21=58
4+5+6+13+14+22=64
As I now have found the pattern, this would allow me to predict the following 3-step stairs if I was still using the method. In order to check if my prediction is right I would have to again add up the numbers in the 3-step stair shape.
5+6+7+14+15+23
= 70
This proves my Prediction to be correct. Therefore I’m able to find the total number of the 3-step stair by just adding 6. However this technique would restrict me from picking out a random 3-step stair shape out of the 9cm by 9cm grid. As I would have to follow the grid side ways from term 1 towards e.g. term 20. Thus making it’s time consuming.
After analysing the grids, the table of results and my prediction I have summed up an algebraic equation which would allow me to find out the total of any 3-step stair shape in a matter of minutes.
The criterion of why I haven’t explained what B stands for is because I haven’t found it yet. Nevertheless using the annotated notes on the formula my formula would look like this now:
Now I would need to find the value of b in order to use my formula in future calculations. Therefore I will take the pattern number of the total:
To conclusion my new formula would be:
8cm by 8cm grid
The Calculations
1+2+3+9+10+17= 42
2+3+4+10+11+18= 48
3+4+5+11+12+19=54
4+5+6+12+13+20=60
As I now have found the pattern, this would allow me to predict the following 3-step stairs if I was still using the method. In order to check if my prediction is right I would have to again add up the numbers in the 3-step stair shape.
5+6+7+13+14+21
= 66
This proves my Prediction to be correct. Therefore I’m able to find the total number of the 3-step stair by just adding 6. However this technique would restrict me from picking out a random 3-step stair shape out of the 9cm by 9cm grid. As I would have to follow the grid side ways from term 1 towards e.g. term 20. Thus making it’s time consuming.
After analysing the grids, the table of results and my prediction I have summed up an algebraic equation which would allow me to find out the total of any 3-step stair shape in a matter of minutes.
The criterion of why I haven’t explained what B stands for is because I haven’t found it yet. Nevertheless using the annotated notes on the formula my formula would look like this now:
Now I would need to find the value of b in order to use my formula in future calculations. Therefore I will take the pattern number of the total:
To conclusion my new formula would be:
11cm by 11cm grid
After analysing the grids, the table of results and my prediction I have summed up an algebraic equation which would allow me to find out the total of any 3-step stair shape in a matter of minutes.
The criterion of why I haven’t explained what B stands for is because I haven’t found it yet. Nevertheless using the annotated notes on the formula my formula would look like this now:
Now I would need to find the value of b in order to use my formula in future calculations. Therefore I will take the pattern number of the total:
To conclusion my new formula would be:
12cm by 12cm grid
- 1+2+3+13+14+25= 58
- 2+3+4+14+15+26= 64
- 3+4+5+15+16+27= 70
- 4+5+6+16+17+28=76
Pattern= +6
As I now have found the pattern, this would allow me to predict the following 3-step stairs if I was still using the method. In order to check if my prediction is right I would have to again add up the numbers in the 3-step stair shape.
5+6+7+14+15+23
= 82
This proves my Prediction to be correct. Therefore I’m able to find the total number of the 3-step stair by just adding 6. However this technique would restrict me from picking out a random 3-step stair shape out of the 9cm by 9cm grid. As I would have to follow the grid side ways from term 1 towards e.g. term 20. Thus making it’s time consuming.
After analysing the grids, the table of results and my prediction I have summed up an algebraic equation which would allow me to find out the total of any 3-step stair shape in a matter of minutes.
The criterion of why I haven’t explained what B stands for is because I haven’t found it yet. Nevertheless using the annotated notes on the formula my formula would look like this now:
Now I would need to find the value of b in order to use my formula in future calculations. Therefore I will take the pattern number of the total:
To conclusion my new formula would be:
After analysing the summary of results and the diagrams above I’ve discovered a link between the difference and the change in formula. As you can see in the highlighted part in the table is the 10cm by 10cm grid. I have discovered the formula for the grid in part one. And looking at the other grids I noticed that when I went one grid down from 10cm by 10cm grid the formula went down by 4. However when I went up a grid the formula went up by four. Therefore I believe the formula of finding a 3-step stairs in different grid you’ll have to. If I’m going down grids then:
However if I’m going up a grid
Firstly the formula when finding the stairs total in the 10cm by 10cm grid is as following: 6n+44
N being the bottom corner number in this case number one. However this formula can’t be used when dealing with other grids. To Conclude I predict the diagram below might be the algebraic formula I’m looking for.
The criterion of why I suggested the algebraic formula might possible is the correct one when dealing with the 10cm by 10cm grid. Because if you add n by 1 it will add up to 2, this is the number on the above stair shape. Also if you add n (which is 1) by 10 it will add up to 11, this is also the number that occurs on the above stair shape. To check if my algebraic formula is correct I will randomly selected another stair shape in the 10cm by 10 cm grid.
As I haven’t discovered my formula yet I will just experiment with the stair shape and see how my formula would look like.
The same method is going to be used when dealing with a 3-step stairs, 4-step stairs and 5-step stairs. However the method changes a little bit when dealing with bigger stairs shapes. As shown below:
As you’ve probably have spotted the higher the stair goes the higher the number. For example on a 2-step stair it’s only N+ (n+1) + (n+10). Whereas with the 3 step stairs, due to the fact it’s a step higher another 10 is added to the number below the 20. As I haven’t discovered my formula yet I will just experiment with the stair shape and see how my formula would look like.
The 4 step stairs are similar to the 3-step stairs only it varies a step thus meaning it will go up by another 10.
I predicted that as the steps go higher the number adds another 10. For example on a 2-step stair it’s only N+ (n+1) + (n+10). Whereas with the 3 step stairs, due to the fact it’s a step higher another 10 is added to the number below the 20, and so going on. As I haven’t discovered my formula yet I will just experiment with the stair shape and see how my formula would look like.
The 5- step stairs are similar to the 4-step stairs only it varies a step thus meaning it will go up by another 10.
I predicted that as the steps go higher the number adds another 10. For example on a 2-step stair it’s only N+ (n+1) + (n+10). Whereas with the 3 step stairs, due to the fact it’s a step higher another 10 is added to the number below the 20, and so going on.
.,..
After analysing the grids and the calculation I notice that first term total in each stair shape have something in common. This is the fact that if you all the N figures times what the value of n is +the multiple of 10 in the stair shape. (These being 10, 20 and 30) You get and outcome number however the with the second equation I’ve notice that the stair total of the stair shape before is the total of the numbers in the stairshape. For example take 4-step stairs
- No. Of N x N+ (add all the multiples of 10 in the stair shape + the stairs total of the stairshape before).
- 10 x n
- N=1
- 10 x1= 10
- 10 + ( 60+ 50 )
- 50 = the total of the 3-step stairs
- 120
To prove my theory is correct; I will have to add the numbers up using the basic method of adding
- 1+2+3+4+11+12+13+21+22+31
However this algebraic formula has two requirements, which would restrict me from using this equation anywhere in the grids;
- You need to know what the total number of the star shape before the stairshape is. For example you need to know the three step stairs 3-step stairs when calculating the total of the 4-step stairs.
- Also the stair shapes need to have the n term is common. For example n is 1 for the 3-step stairs and also for the 4-step stairs Otherwise it wont work
I have discovered 3 patterns:
-
The first pattern is as following; the beginning first digit or digits of the formula e.g. (3+ (10+1)) goes up by 3 then 4 then 5. For example 2-step’s formula is 3+ (10+1) and it goes up by 3 because 3-step formula is 6+ (30+14) et cetera. Therefore I predict the next first digit/ digits of the formula will go up by 6,
- The second pattern is similar however it’s to do with the multiples of ten in the stairshape it goes up by 20 in the first formula then by 30 in the next and 40 in the 5-step formula. Therefore I predict the 6-step formula to go up by 50.
- The third pattern is that the last digit or digits are the total of the stairshape before (shown in analyse.)
To see if my prediction is correct I will have to add all the number in the 6-step stair shape.
1+2+3+4+5+6+11+12+13+14+15+21+22+23+24+31+32+33+41+42+51
To ensure my result is accurate I will use a calculator for the long multiplication show above.
= 406
This proves my prediction correct. However now I will have to find an algebraic formula to calculate any stairshape in any grid. I wouldn’t have to the 10cm by 10cm grid or the 3-step stairs as I already done that. Therefore I would only experiment with the 8cm by 8cm grid, 9cm by 9cm grid, 11 by 11 cm grid and the 12cm by 12cm grid. In all those grids I would only do the 2-step, 4-step and 5 step stair shape.
8cm by 8cm grid- 2-step stair
Calculation
8cm by 8cm grid- 4-step stair
8cm by 8cm grid- 4-step stair
9cm by 9cm grid
1+2+10=13
2+3+11=16
3+4+12=19
9cm by 9cm grid- 4-step stairs
The Calculations
1+2+3++410+11+12+19+20+28= 110
2+3+4+5+11+12+13+20+21+29= 120
3+4+5+6+12+13+14+21+22+30=130
9cm by 9cm grid- 5-step stairs
The Calculations
1+2+3+4+5+10+11+12+13+19+20+21+28+29+37= 215
2+3+4+5+6+11+12+13+14+20+21+22+29+30+38= 230
3+4+5+6+7+12+13+14+15+21+22+23+30+31+39=245
11cm by 11cm grid
11cm by 11cm grid
11cm by 11cm grid
I have discovered 3 patterns:
- When going up a step; the nth term will go from e.g. 4-step formula 10n+90 to 15n=180. Therefore the numbers are going up the same way I’ve demonstrated in the table of result with the 10cm by 10cm grid. When the numbers go from +3, +4, +5, +6 et cetera.
- In every grid when going up a step the formula’s end digits double up. Shown in the table above. However I excluded 3-step formula and the 10cm by 10cm grid. Therefore it looks like the first formula in the grid is times and the double. For example look at the 9cm by 9cm grid going downwards. 12 x 10= 120 +120= 240.
- Lastly, when going across the table the steps formula first digits remain the same however it a goes up by +1, +2 +3 and so on.
To conclude I have now found and demonstrated the relation ship between the stairs total and the stairs shape. However From this piece of coursework I have found out what number stairs are and I have investigated the relationship between their totals and positions to work out a formula. This led me on to investigating the relationships between totals and positions when the stair size and grid size is different.