Maths tomato investigation

The best thing about the above expression is that it works for any tomato going bad first in any sized square.

Now, letīs have a look at something slightly more complex, an infinitely big tray of tomatoes. If a tomato goes bad in the corner of an infinitely big tray of tomatoes, a triangular number type pattern will happen. I will illustrate it in the table on the next page:

: Hour (H) No. bad toms bad afterhour H (N) No. toms going bad at hour H (B)

0 1 1 1 1+2 2 2 1+2+3 3 3 1+2+3+4 4 4 1+2+3+4+5 5 5 etc.

These are called triangular numbers because as you write them down they form a triangle type shape. For any infinitely big square, an expression can be formed to show the growth rate of the amount of tomatoes going bad each hour. In this case itīs H + 1 = B . This is because at hour 0 the first tomato went bad, therefore immediately the expression works. Next, because itīs a triangular number pattern, the amount going bad each hour will increase by one each time:

A table can also be drawn which relates the hour at which the tomatoes go bad with the total number of bad tomatoes:

Hour (H) Total number of bad tomatoes (N)

0 1 1 3 2 6 3 10 4 15 5 21 From this table an expression can be used to relate H to N. It contains a new variable which Iīll call P. P = H+1. Therefore the expression is P2 + P = N . This means that if we take three as an example, 2 3+1 = 4, squared, plus 4 equals 20, divided by two equals 10. This works because the second difference between the number of bad tomatoes is constant(as illustrated in the table on the next page):

N difference 1 difference 2

1 3 +2 6 +3 +1 10 +4 +1 15 +5 +1 21 +6 +1

It is always +1, this means that an expression to describe it will envolve H2, or in this case P2. Clearly though ...