The problem asks us to find the general term $a_n$ of a sequence defined by the recurrence relation $a_{n+1} = 3a_n + 4$ with the initial condition $a_1 = 2$.

1. Problem Description

The problem asks us to find the general term $a_n$ of a sequence defined by the recurrence relation $a_{n+1} = 3a_n + 4$ with the initial condition $a_1 = 2$.

2. Solution Steps

First, let's rewrite the recurrence relation in the form $a_{n+1} + \alpha = 3(a_n + \alpha)$. Expanding the right side, we get $a_{n+1} + \alpha = 3a_n + 3\alpha$. Comparing this to $a_{n+1} = 3a_n + 4$, we have $\alpha = 3\alpha - 4$, which means $2\alpha = 4$, so $\alpha = 2$.

Therefore, the recurrence relation can be written as $a_{n+1} + 2 = 3(a_n + 2)$. Let $b_n = a_n + 2$. Then the recurrence relation becomes $b_{n+1} = 3b_n$. This is a geometric sequence with common ratio $3$. We have $b_1 = a_1 + 2 = 2 + 2 = 4$. The general term for a geometric sequence is given by

$b_n = b_1 r^{n-1}$,

where $b_1$ is the first term and $r$ is the common ratio. In this case, $b_1 = 4$ and $r = 3$, so

$b_n = 4 \cdot 3^{n-1}$.

Since $b_n = a_n + 2$, we have $a_n = b_n - 2 = 4 \cdot 3^{n-1} - 2$.

3. Final Answer

$a_n = 4 \cdot 3^{n-1} - 2$