Answer
Prove that $p(1)$ is true.
Prove that $p(k)$ true implies $p(k+1)$ true.
Work Step by Step
Let's note by $p(n)$ the property we want to prove:
$$p(n):\text{ } a_n=2\cdot 3^n-2.$$
The sequence is defined by the recursive rule:
$$\begin{cases}
a_1=4\\
a_{n+1}=3a_n+4.
\end{cases}$$
$\textbf{Step 1:}$ Prove $p(1)$ is true.
Set $n=1$ and check if the property is verified:
$$\begin{align*}
a_1&\stackrel{?}{=}2\cdot 3^1-2\\
4&=4\checkmark.
\end{align*}$$
Therefore $p(1)$ is true.
$\textbf{Step 2:}$ Prove that if $p(k)$ is true, then $p(k+1)$ is true.
We assumed $p(k)$ is true:
$$p(k):\text{ } a_k=2\cdot 3^k-2.\tag1$$
We have to prove:
$$p(k+1):\text{ } a_{k+1}=2\cdot 3^{k+1}-2.\tag2$$
Rewrite $a_{k+1}$ using the sequence's rule and $a_{k}=2\cdot 3^k-2$:
$$\begin{align*}
a_{k+1}&=3a_k+4\\
&=3(2\cdot 3^k-2)+4\\
&=2\cdot 3\cdot 3^k-6+4\\
&=2\cdot 3^{k+1}-2.
\end{align*}$$
We got Eq. $(2)$, therefore if $p(k)$ is true, then $p(k+1)$ is also true.
Both steps being proved, the given formula is true for any natural number $n$.
