Refer this link for solving recurrence relations(a very good read): Solving recurrences
The question is in the form of second-order recurrence relation : $C_{0}x_{r}+C_{1}x_{r-1}+C_{2}x_{r-2}=0$
The characteristic equation of the recurrence is given by: $C_{0}m^{2}+C_{1}m+C_{2}=0$ (refer the link above for detailed explanation).
$\therefore$ The given recurrence can be written as : $a_{r}-a_{r-1}-2a_{r-2}=0$
$\Rightarrow$ $m^{2}-m-2=0$ $\Rightarrow$ $(m+1)(m-2)=0$ $\Rightarrow$ $m_{1}=-1$ and $m_{2}=2$
For real and distinct roots of the quadratic equation, the solution to the recurrence is given by : $x_{r}=c_{1}m_{1}^{r}+c_{2}m_{2}^{r}$
Using the given data of $T_{0}$ and $T_{1}$, we get $c_{1}(-1)^{0}+c_{2}(2)^{0}=2$ and $c_{1}(-1)^{1}+c_{2}(2)^{1}=7$
$\Rightarrow$ $c_{1}=-1$ and $c_{2}=3$
$\therefore$ Solution to the recurrence would be: $a_{r}=3(2)^{r}-(-1)^{r}$
Option D is correct.
