If the population of a town is p in the beginning of any year then it becomes 3+2p in the beginning of the next year. If the population in the beginning of 2019 is 1000, then the population in the beginning of 2034 will be

Correct Option: 3 (representing $2^{15}\left(1003\right)-3$)

Let $P_n$ be the population at the beginning of year $n$ (where $n=0$ corresponds to 2019).
We are given the recurrence relation:
$P_{n+1}=2P_n+3$

Let us write down the terms for the first few years to find a pattern:
$P_0=1000$ (beginning of 2019)
$P_1=2\left(1000\right)+3$ (beginning of 2020)
$P_2=2P_1+3=2\left(2\right(1000)+3)+3=2^2\left(1000\right)+2\left(3\right)+3$
$P_3=2P_2+3=2^3\left(1000\right)+2^2\left(3\right)+2\left(3\right)+3$

In general, for the beginning of year $2019+t$ (which is $P_t$):
$P_t=2^t\left(1000\right)+3(2^{t-1}+2^{t-2}+\dots +2+1)$

The expression in the parentheses is a geometric progression with first term $a=1$, common ratio $r=2$, and $t$ terms.
The sum of this GP is:
$\frac{1(2^t-1)}{2-1}=2^t-1$

Substituting this back into the formula:
$P_t=2^t\left(1000\right)+3(2^t-1)$
$P_t=2^t\left(1000\right)+3·2^t-3$
$P_t=2^t(1000+3)-3$

We need the population in the beginning of 2034.
Here, $t=2034-2019=15$ years.
Substituting $t=15$:
$P_{15}=2^{15}\left(1003\right)-3$