A lake contains 200,000m^3$ of water, 7% of which is chemical pollution. Every day 2000m^3$ of polluted water flows out of the lake.
(a) Suppose that every day $2000m^3$ of clean water flows into the lake. If $P_n$ denotes the volume of pollutant in the lake on day $n$, show that $P_{n+1}= $0.99P_n$. Hence show that it will be 194 complete days before the pollution drops to the safe level deemed to be 1%.
I've shown that $P_{n+1}=0.99P_n$ as every day out of the $2000m^3$ of water that flows out of the lake each $m^3$ contains $P_n/200,000m^3$ of pollutant So on day $n+1$,
$P_{n+1}=P_n -2000P_n/200,000$
leading to
$P_{n+1}=0.99P_n$
I also know initially on day $0$ the lake contains
$P_0=0.07 \cdot 200,000=14,000m^3$ of polluted water
I have the solutions to this question and it says solving recursively
$P_n=0.99^nP_0$
which obviously leads to
$2000=0.99^n\cdot 14,000$
I know the 2000 comes from needing the pollution level to be within 1% of the total volume of water and taking logs in this last equation will lead to $nlog0.99=-\log7$ then $n=193.62$ showing that it will be 194 complete days before the pollution drops to the safe level.
My main confusion is solving recursively from $P_{n+1}=0.99P_n$ to $P_n=0.99^nP_0$ I really don't get how one solves this if someone could help it would be greatly appreciated!
We know the relation $P_{n+1} = 0.99P_n$. But we also know that $P_n = 0.99P_{n-1}$, etc... Now we substitute and find $$ P_{n} = 0.99P_{n-1} = 0.99 (0.99 P_{n-2}) = 0.99 (0.99 (0.99P_{n-3})) = 0.99^{k}P_{n-k}$$ Thus we conclude that $P_n = 0.99^n P_0$ (just plug in $n=k$).