6. 7A + B = 5
I then needed to take equation 5. away from equation 6. so that B would be cancelled out and I would be left with A.
6. 7A + B = 5
- 5. 3A + B = 15
4A = 2
I have now worked out what 4A equals so if I divide 2 by 4 my answer is 0.5, this shows that A = 0.5. I shall now substitute 0.5 in my equations to work out what B and represents, I already know that C represents 1 because it is the 1st number in the nth term.
I will substitute these numbers to see if the formula works:
- 0.5 + B + 1 = 3
By substituting A an C in equation 1. I can work out that B represents 1.5
- 2 + 3 + 1 = 6
- 4.5 + 4.5 + 1 = 10
- 8 + 6 + 1 = 15
Natalie Hayes 10E Group 3 Page 3
I have substituted the numbers into all of the equations and it works for all of them.
A = 0.5
B = 1.5
C = 1
Now I know what A, B and C stand for I can now substitute these into the formula:
0.5 n² + 1.5n + 1
Side
I am now going to investigate how many the total amounts of bad tomatoes after a certain amount of hours, with the bad tomato starting at the side in a box with unlimited sides.
Here is my table of results after looking at the diagram on page 6.
By looking at this table of results, I can now produce a sequence.
1 2 3 4 5
4 9 16 25 36
+3 +5 +7 +9
+2 +2 +2
The first thing I noticed about this sequence is that the numbers in the first differnce are all odd numbers. As you can see this sequence it is similar to the one I investigated before, as second differences are also present. Again, I will have to use the formula An² + Bn + C to work out an nth term. I will do this again by working out the simultaneous equations to find out what A, B and C represent.
1. A + B + C = 4
2. 4A + 2B + C = 9
3. 9A + 3B + C = 16
4. 16A + 4B + C = 25
Natalie Hayes 10E Group 3 Page 4
2. 4A + 2B + C = 9
- 1. A + B + C = 4
5. 3A + B = 5
4. 16A + 4B + C = 25
- 3. 9A + 3B + C = 16
6. 7A + B = 9
I have done the same as I did with the 1st investigation, so I will continue by subtracting equation 5. from 6.
6. 7A + B = 9
- 5. 3A + B = 5
4A = 4
4 divided by 1 = 1 therefore A = 1
I have now worked out what A equals so I now need to substitute 1 in my equations to work out what B represents, as C represents 1 again as it is the first number in the nth term.
- 1 + B + 1 = 4
By substituting A and C in equation 1. I can work out that B
comes to the total amount of 2. I will now substitute all these numbers into the equation to check if they are correct.
- 4 + 4 + 1 = 9
- 9 + 6 + 1 = 16
- 16 + 8 + 1 = 25
All of these equations add up correctly so the numbers I have found are all correct.
A = 1
B = 2
C = 1
It is now possible for me to substitute these numbers into the formula:
n² + 2n + 1
Natalie Hayes 10E Group 3 Page 5
Middle
I found this task the most complex as it took awhile for me to work out exactly how many tomatoes were bad after a certain amount of hours, but once I got the hang of it I then found it more simple as I could follow the same pattern as before, here is my working:
1 2 3 4 5
5 13 25 41 61
+8 +12 +16 +20
+4 +4 +4
I noticed that all the numbers in the 1st differences are they are all even numbers. As you can see this sequence is slightly different, as the second difference is of a different quantity (4) even though this is different the method still stays the same.
1. A + B + C = 5
2. 4A + 2B + C = 13
3. 9A + 3B + C = 25
- 16A + 4B + C = 41
2. 4A + 2B + C = 13
- 1. A + B + C = 5
5. 3A + B = 8
4. 16A + 4B + C = 41
- 3. 9A + 3B + C = 25
6. 7A + B = 16
6. 7A + B = 16
- 5. 3A + B = 8
4A = 8
8 divided by 4 = 2 therefore A = 2 ans C = 1
Natalie Hayes 10E Group 3 Page 6
I then substituted A with 2 in all my equations:
- 2 + B + 1 = 5
from looking at this equation it becomes apparent that B = 2
- 8 + 4 + 1 = 13
- 18 + 6 + 1 = 25
- 32 + 8 + 1 = 41
A = 2
B = 2
C = 1
2n² + 2n + 1
Middle
By enlarging the diagrams I can clearly see
How many tomatoes go bad after how many
hours, without having to work it all out.
Corner
Natalie Hayes 10E Group 3 Page 7
Side
After thoroughly looking at the diagrams I noticed a pattern in the tray that has the bad tomato starting in the corner, I noticed that it uses triangular numbers.
I decided to investigate and see if I could use the triangular numbers formula to produce my own formula for an nth term. A number n is said to be a "Triangle Number" if it can arrange n number of objects into a solid triangle pattern: (1,3,6 and 10 are triangle numbers)
The nth triangle number can be found with n = ½n (n + 1)
Natalie Hayes 10E Group 3 Page 8
Therefore, the 3rd triangle number is = ½ 3 (3 + 1) = 6
If I were to use a tray with limited sides eg a tray 8 by 8 then I would get slighty different results:
As you can see from the diagram, after the row of tomatoes touching corner to corner in a diagonal (the longest row of bad tomatoes) the row of tomatoes start to decrease. The formula for triangle numbers ½n (n + 1) doesn’t apply to this pattern any longer, but it continues to work in a box with endless sides.
Now that I have found formulas to show how many tomatoes go bad in a tray with unlimited sides, I will now attempt to find out if there is a pattern when you use different size trays e.g. 4 by 4, 3 by 3.
Natalie Hayes 10E Group 3 Page 9
Here are some diagrams to show how many tomatoes go bad in a tray of 4 by 4, 3 by 3, and 2 by 2, with the bad tomato starting in the corner (1).
By looking at these diagrams, I can see a pattern begin to form.
1 2 3 4 5
0 2 4 6 8
+2 +2 +2 +2
There is only 1st differences present this time so I will use simultaneous equations to work out a formula to find the nth term.
- A + B = 0
- 2A + B = 2
- 3A + B = 4
- 4A + B = 6
- 2A + B = 2
- 1. A + B = 0
A = 2
If I substitute 2 for A in equation 2. then I will get 4 + B = 2
Natalie Hayes 10E Group 3 Page 10
Therefore B = -2
Nth term = 2n - 2
Side
Here are a few diagrams that I used to help me make a table of results to show how many tomatoes go bad in different size trays, with the bad tomato starting at the side (5). My first box had to be at least 3 by 3 otherwise the starting tomato would not be touching 3 other tomatoes so the results would be incorrect.
3 4 5 6
3 5 7 9
+2 +2 +2
There is also 1st differences present so again I will use simultaneous equations to work out a formula to find the nth term.
1. A + B = 3
2. 2A + B = 5
3. 3A + B = 7
- 4A + B = 9
- 2A + B = 5
- 1. A + B = 3
A = 2
Natalie Hayes 10E Group 3 Page 11
If I substitute 2 for A in equation 2. then I will get 4 + B = 5
Therefore B = 1
Nth term = 2n + 1
Corner
When investigating if the figures change in different size trays when the starting bad tomato is positioned in the middle I realised that I could only use trays with odd sides e.g. 3 by 3, 5 by 5. as trays with even sides do not have an exact centre position. Here are my diagrams that I used and a table to show my results.
Natalie Hayes 10E Group 3 Page 12
3 5 7 9
2 4 6 8
+2 +2 +2
I can see by looking at my results that the nth term is n - 1, for example it would take 10 hours for the tomatoes to go bad in a box of 11 by 11.
Conclusion
For part 1 of this investigation I have found out how the tomatoes go bad in trays of different sizes and trays of unlimited sides, I have noticed differences and patterns if you change the position of the starting bad tomato. After researching and using my knowledge I have been able to find many formula’s so it is now possible to understand how the tomatoes in the tray go bad.