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

1. Problem Description

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

2. Solution Steps

Let's try to transform the recursive relation into a more manageable form.

Assume that $a_{n+1} + x = 3(a_n + x)$ for some constant $x$. Then $a_{n+1} = 3a_n + 3x - x$, so $a_{n+1} = 3a_n + 2x$.

Comparing this with $a_{n+1} = 3a_n + 4$, we have $2x = 4$, so $x = 2$.

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

Let $b_n = a_n + 2$. Then $b_{n+1} = 3b_n$, which means that $\{b_n\}$ is a geometric sequence with a common ratio of $3$.

We have $b_1 = a_1 + 2 = 2 + 2 = 4$.

The general term for $b_n$ is given by $b_n = b_1 \cdot 3^{n-1}$.

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

Since $b_n = a_n + 2$, we have $a_n = b_n - 2$.

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

We can check the first few terms.

$a_1 = 4 \cdot 3^{1-1} - 2 = 4 \cdot 3^0 - 2 = 4 \cdot 1 - 2 = 2$

$a_2 = 3a_1 + 4 = 3(2) + 4 = 10$. Also, $a_2 = 4 \cdot 3^{2-1} - 2 = 4 \cdot 3 - 2 = 12 - 2 = 10$.

$a_3 = 3a_2 + 4 = 3(10) + 4 = 34$. Also, $a_3 = 4 \cdot 3^{3-1} - 2 = 4 \cdot 9 - 2 = 36 - 2 = 34$.

3. Final Answer

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