Solution – Recurrence $f(n)=3f(n/2)+6$, $f(1)=1$ by Successive Substitution

Assume $n=2^k$ ($k=\log_2 n$).

Unfold:
$$\begin{aligned} f(n) &= 3f(n/2)+6 \\ &= 3[3f(n/4)+6]+6 = 3^2 f(n/2^2)+6\cdot3+6 \\ &= 3^2[3f(n/2^3)+6]+6\cdot3+6 = 3^3 f(n/2^3)+6\cdot3^2+6\cdot3+6\\ &\;\vdots \\ f(n) &= 3^k f(n/2^k)+6\sum_{i=0}^{k-1}3^i \end{aligned}$$

Since $n/2^k=1$, $f(n/2^k)=f(1)=1$ and $\sum_{i=0}^{k-1}3^i = (3^k-1)/(3-1)=(3^k-1)/2$.

$$f(n)=3^k \cdot 1 +6\cdot\frac{3^k-1}{2}=3^k+3(3^k-1)=4\cdot3^k-3$$

With $k=\log_2 n$:
$$\boxed{f(n)=4\cdot 3^{\log_2 n}-3 = 4\cdot n^{\log_2 3}-3}$$ for $n$ power of 2.

Asymptotics:
$$f(n)=\Theta(n^{\log_2 3}) = \Theta(n^{1.58496\dots})$$

Check by Master Theorem: $a=3$, $b=2$, $n^{\log_b a}=n^{\log_2 3}$ dominates $f(n)=\Theta(1)$ case 1 → $\Theta(n^{\log_2 3})$ consistent.

For general $n$ (not power of 2), embed between powers of 2 → same $\Theta$.