$\tt T(n) = 9T(\frac{n}{2}) + n$
Here $a = 9$, $b = 2$, $k = 1$. and $a \gt b^k$
Thus, $\color{navy}{ T(n) = \theta(n^{log_2(9)})}$, only $3^n$ upper bounds answer, but it isn't average bound. So, it's wrong to say $T(n) = \theta(n^{log_2(9)}) = \theta(3^n)$
If recurrence was $\tt T(n) = 9T(\frac{n}{3}) + n$, Then correct answer would be $\theta(n^2)$
