We are given a sequence $a_n$ such that $a_1 = 2$ and $a_{n+1} = 3a_n + 4$. We need to find the general term $a_n$.

1. Problem Description

We are given a sequence $a_n$ such that $a_1 = 2$ and $a_{n+1} = 3a_n + 4$. We need to find the general term $a_n$.

2. Solution Steps

First, let's rewrite the recurrence relation as

$a_{n+1} + x = 3(a_n + x)$.

$a_{n+1} = 3a_n + 2x$.

Comparing with the original relation $a_{n+1} = 3a_n + 4$, we have $2x = 4$, which gives $x = 2$.

Therefore, $a_{n+1} + 2 = 3(a_n + 2)$.

Let $b_n = a_n + 2$. Then $b_{n+1} = 3b_n$.

This means that the sequence $b_n$ is a geometric sequence with common ratio

3. Since $b_1 = a_1 + 2 = 2 + 2 = 4$, we have $b_n = b_1 \cdot 3^{n-1} = 4 \cdot 3^{n-1}$.

Therefore, $a_n + 2 = 4 \cdot 3^{n-1}$, which gives $a_n = 4 \cdot 3^{n-1} - 2$.

3. Final Answer

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