I’m struggling with understanding how to solve part B of the following problem:
Consider an outdoor pool initially filled with 20,000 gallons of water. Each day, 4% of the water in the pool evaporates. Each morning at 10:00am, W gallons of water are added back to the pool where W is a constant.
a) Let $A_n$ be the number of gallons of water in the pool immediately after water is added back into the pool for the nth time. Given that $A_1 = 19200 + W$, find $A_2$ and $A_3$
I have solved part A, and according to the answer sheet this is correct:
$A_2=(19200+W)(0.96)$, $A_3=((19200+W)(0.96)+W)(0.96)$
b) Find a closed-form expression for $A_n$ (i.e. evaluate any sums and solve any recursion). Note an answer may contain the constant W.
I am having trouble turning my answers from part A into a closed-form expression. I know that the sum of a finite geometric series is equal to $\frac{a(1-r^n)}{(1-r)}$ where $a$ is the initial term and $r$ is the ratio. In this case, it seems like 0.96 would be r. But because the expression for the first term depends on where you start in the summation (recursive), I’m not sure how to proceed.
According to my answer sheet, the answer is $A_n=20,0000(\frac{24}{25})^n+25W(1-(\frac{24}{25})^n$, but I am having trouble understanding why because the answer sheet does not explain it. Can you please help?
I'm going to give an answer based on what I think should be right, as I stated in the comments.
There is a pattern. For ease and clarity I'll take $0.96=k$
$A_0=20000$
$A_1=20000k^1+W(1)$
$A_2=20000k^2+W(k+1)$
$A_3=20000k^3+W(k^2+k+1)$
$\vdots$
$A_n=20000k^n+W(\color{blue}?)$
Let's fill in the question mark.
What we have is a geometric series where the first term is $1$ and the constant ratio $k$. The sum for the first $n$ terms in this series would be:
$\dfrac{1-k^n}{1-k}$
The whole thing is thus:
$A_n=20000(0.96)^n+W\left(\dfrac{1-(0.96)^n}{\dfrac{1}{25}} \right)$
However you choose to write it, the final answer should be:
$\boxed{A_n=20000 \cdot 0.96^n+25W(1-0.96^n)}$
Q.E.D