Given that, the population of a town is $\text{‘P’}$ at the beginning of any year, then the population becomes $\text{‘3+2P’}$ at the beginning of the next year. And the population at the beginning of $2019$ is $1000$.
Now, we can write each year's population in the following manner.
- $\text{Year – 0:}$ Beginning of $ 2019 \longrightarrow \text{P} = 1000 = (1003)2^{0} – 3$
- $\text{Year – 1:}$ Beginning of $ 2020 \longrightarrow 3+2\text{P} = 3+2\times1000 = 2003 = (1003)2^{1} – 3$
- $\text{Year – 2:}$ Beginning of $ 2021 \longrightarrow 3+2\text{P} = 3+2\times2003 = 3 + 4006 = 4009 = (1003)2^{2} – 3$
- $\text{Year – 3:}$ Beginning of $ 2022 \longrightarrow 3+2\text{P} = 3+ 2 \times 4009 = 3+8018 = 8021 = (1003)2^{3}-3$
- $\vdots \;\; \vdots \;\; \vdots$
- $\text{Year – n: } \boxed {(1003)2^{n}-3}, \; n= \text{number of years}$
$\therefore\; \text{Year – 15:}$ Beginning of $ 2034 \longrightarrow (1003)2^{n} – 3 = (1003)2^{15} – 3$
Correct Answer $: \text{C}$