Answer
a) $1.08^nP_{0}-3750(1.08^n-1)$
b) $6898$ catfish
Work Step by Step
Let's note by $P_n$ the catfish population after $n$ months.
a) At the end of the first month the population is:
$$5,000+0.08(5000)-300=1.08(5000)-300.$$
At the end of the second month the population is:
$$1.08(1.08(5000)-300)-300=1.08^2(5000)-1.08(300)-300.$$
At the end of the $n$th month the population is the population from the previous month, increased by $8\%$ from which we subtract the harvested part of $500$:
$$P_n=1.08P_{n-1}-500.$$
b) We determine $P_n$:
$$\begin{align*}
P_n&=1.08P_{n-1}-300\\
&=1.08(1.08P_{n-2}-300)-300\\
&=1.08^2P_{n-2}-1.08(300)-300\\
&=1.08^2(1.08P_{n-3}-300)-1.08(300)-300\\
&=1.08^3P_{n-3}-1.08^2(300)-1.08(300)-300\\
&=.......\\
&=1.08^nP_{0}-1.08^{n-1}(300)-1.08^{n-2}(300)-\dots -300\\
&=1.08^nP_{0}-300(1.08^{n-1}+1.08^{n-2}+\dots+1)\\
&=1.08^nP_{0}-300\cdot\dfrac{1.08^n-1}{1.08-1}\\
&=1.08^nP_{0}-3750(1.08^n-1).
\end{align*}$$
We calculate $A_{12}$:
$$A_{12}=1.08^{12}(5,000)-3750(1.08^{12}-1)\approx 6898.$$
