Solving the recurrence T(n) = 3T(n-2) with iterative method
Note that $3^{n/2-1/2} = \frac{1}{\sqrt{3}} 3^{n/2} = \frac 1{\sqrt{3}} \sqrt{3^n}$. So the $-\frac 12$ indeed becomes a constant factor that is absorbed by the $O()$, but $\frac n2$ in the exponent changes the base and changing the base changes the $O$-class.

The correct answer thus is $T(n) = O(3^{n/2}) = O(\sqrt{3^n})$.