Answer:
F(n) = 225 +(n)300
Explanation:
It is given that the fish counts of last 3 weeks are as follows,
225, 525 and 825.
The difference between consecutive terms is 300.
If we carefully look, these are numbers in arithmetic progression, with first term as 225 and common difference as 300.
Thus, at the end of nth week, with initial amount of fish at n=0 as 225, the fish count is given by,
F(n) = 225 +(n)300, where F(n) in function of total number of fishes and n is the number of weeks.
We can verify as, at n=1, F(1) = 225 + 300 = 525 and F(2) = 225 + 600 = 825, which are already correctly given.