Determine a closed form for this sequence

Question

Every year, 38 % of the amount of fish in a pond die. The 1st of May 2011 there were 5200 fish in the pond. Every year after May 1st 2011, 1900 new fish are added to the pond.

Let $a_n$ be the amount of fish in the pond (directly after new fish have been added) $n$ years after May 1st 2011.

Determine a closed form for this sequence.

How to solve this problem?

My attempt is: $a_n = 5200 - 0,38a_n + 1900n \Rightarrow 1,38a_n = 5200 + 1900n \Rightarrow a_n = \frac{5200 + 1900n}{1,38}$

Answer 1

Here's how I would start: $$\begin{align*}a_0 &= 5200 \\ a_1 &= 5200\cdot 0.62+1900 \\ a_2 &= (5200\cdot 0.62+1900)\cdot 0.62 +1900 = 5200\cdot 0.62^2+1900(1+0.62) \\ a_3 &= 5200\cdot 0.62^3+1900(1+0.62+0.62^2) \\ \vdots\end{align*}$$ Can you find a general $a_n$ from here? A fun bonus question is: If the process repeats indefinitely, how many fish will be left at the end?

Answer 2

hints Looks like $a_{n+1} = (1-0.38) a_n + 1900 = 0.62 a_n + 1900$ with $a_0 = 5200$.

In general you should get $a_n = A 0.62^n + B + Cn$, note that $5200 = a_0 = A + B$ and get other initial conditions and find B and C.

Answer 3

Nope, but it is a good start, $a_1=5200+1900$,$a_2=0.68^2\cdot 5200+1900\cdot 0.68+1900$,$a_3=0.68^3\cdot 5200+1900\cdot 0.68^2+1900\cdot 0.68+1900$, see the pattern?

$$a_n=5200\cdot 0.68^n+1900\cdot\frac{1-0.68^{n-1}}{0.32}$$