We are given a recursive sequence defined by $a_1 = 2$ and $a_{n+1} = 3a_n + 4$. We want to find the general term $a_n$ of the sequence.

1. Problem Description

We are given a recursive sequence defined by $a_1 = 2$ and $a_{n+1} = 3a_n + 4$. We want to find the general term $a_n$ of the sequence.

2. Solution Steps

We have the recursive relation $a_{n+1} = 3a_n + 4$. We can rewrite this as $a_{n+1} + c = 3(a_n + c)$ for some constant $c$. Expanding the right side, we have $a_{n+1} + c = 3a_n + 3c$. Comparing this with $a_{n+1} = 3a_n + 4$, we have $3c = c + 4$, so $2c = 4$ and $c = 2$.

Therefore, $a_{n+1} + 2 = 3(a_n + 2)$. Let $b_n = a_n + 2$. Then $b_{n+1} = 3b_n$, which means $b_n$ is a geometric sequence with common ratio

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

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

We can check the first few terms:

$a_1 = 4 \cdot 3^{1-1} - 2 = 4 \cdot 1 - 2 = 2$.

$a_2 = 4 \cdot 3^{2-1} - 2 = 4 \cdot 3 - 2 = 10$.

Using the recursive relation, $a_2 = 3a_1 + 4 = 3(2) + 4 = 10$.

$a_3 = 4 \cdot 3^{3-1} - 2 = 4 \cdot 9 - 2 = 34$.

$a_3 = 3a_2 + 4 = 3(10) + 4 = 34$.

3. Final Answer

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