We are given a sequence $\{a_n\}$ defined by the recurrence relation $a_{n+1} = 3a_n + 4$ with the initial condition $a_1 = 2$. We need to find the general term $a_n$ of this sequence.

1. Problem Description

We are given a sequence $\{a_n\}$ defined by the recurrence relation $a_{n+1} = 3a_n + 4$ with the initial condition $a_1 = 2$. We need to find the general term $a_n$ of this sequence.

2. Solution Steps

First, we try to find a constant $c$ such that $a_{n+1} + c = 3(a_n + c)$. This gives us $a_{n+1} = 3a_n + 2c$. Comparing this with the given recurrence relation $a_{n+1} = 3a_n + 4$, we have $2c = 4$, which implies $c = 2$.

Now, let $b_n = a_n + 2$. Then, $b_{n+1} = a_{n+1} + 2 = 3a_n + 4 + 2 = 3a_n + 6 = 3(a_n + 2) = 3b_n$. This means that the sequence $\{b_n\}$ is a geometric sequence with common ratio

3. Also, $b_1 = a_1 + 2 = 2 + 2 = 4$.

Therefore, $b_n = b_1 \cdot 3^{n-1} = 4 \cdot 3^{n-1}$.

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

3. Final Answer

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