Fire Ant mounds are rising in my yard: if a mound can have only one queen, and she lays $50$ viable eggs per day, and if $2\%$ of the population dies per day, what is the limit of population growth? Obviously, when the death rate equals $50$ per day, no further growth can occur. My solution attempt:
- population on day 1: $0.98*50$
- population on day 2: $0.98*50 + 0.98^2*50$ $\quad$ (new hatchlings + survivors from day 1)
- population on day 3: $0.98*50 + 0.98^3*50 + 0.98^2*50$
- Generalizing: $P = \sum_{t=1}^{?} 0.98^t*50$
A recursion in this form $$ x_0 = k\\ x_{n+1} = p + \alpha x_n $$ where $\alpha\neq0$, is solved by the explicit formula
$x_n = \frac{1-\alpha^n}{1-\alpha}p+\alpha^nk$
If, moreover, $-1<\alpha<1$, there exists the asymptotic limit
$\lim_{n\rightarrow+\infty} x_n= \lim_{n\rightarrow+\infty}(\frac{1-\alpha^n}{1-\alpha}p+\alpha^nk)= \frac{p}{1-\alpha}$
In your problem, $x_n$ represents the number of non-queen ants on the n-th day.
The mathematisation you ask for has $\alpha=0.98,\ k=0,\ p=0.98*50=49$, and the formula tells you that your colony asymptotically attends $2450$ individuals, excluding the queen.
P.S.: If, on the other hand, you wanted to assume Gregory Grant's point of view (which makes more sense to me, but whatever), you should put $\alpha=0.98,\ k=0,\ p=50$ and run the same calculation, obtaining $2500$ instead of $2450$.