We are given a sequence $\{a_n\}$ defined recursively by $a_1 = 2$ and $a_{n+1} = 3a_n + 4$. The goal is to find a formula for the general term $a_n$ of this sequence.

1. Problem Description

We are given a sequence $\{a_n\}$ defined recursively by $a_1 = 2$ and $a_{n+1} = 3a_n + 4$. The goal is to find a formula for the general term $a_n$ of this sequence.

2. Solution Steps

First, let's find a particular solution to the recursion $a_{n+1} = 3a_n + 4$. Assume that the sequence converges to a constant value $L$. Then, in the limit as $n$ goes to infinity, we have $L = 3L + 4$. Solving for $L$, we get $L = -2$.

Now, let's define a new sequence $b_n = a_n - L = a_n + 2$. Then $a_n = b_n - 2$.

Substituting this into the original recurrence, we get

$b_{n+1} - 2 = 3(b_n - 2) + 4$,

$b_{n+1} - 2 = 3b_n - 6 + 4$,

$b_{n+1} = 3b_n$.

This means that $b_n$ is a geometric sequence with common ratio $3$.

We have $b_n = b_1 \cdot 3^{n-1}$.

Since $b_n = a_n + 2$, we have $b_1 = a_1 + 2 = 2 + 2 = 4$.

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

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

Let us verify this for a few values of $n$.

For $n=1$, we have $a_1 = 4 \cdot 3^{1-1} - 2 = 4 \cdot 3^0 - 2 = 4 - 2 = 2$. This matches the given initial condition.

For $n=2$, we have $a_2 = 3a_1 + 4 = 3(2) + 4 = 6+4=10$.

Using the formula, $a_2 = 4 \cdot 3^{2-1} - 2 = 4 \cdot 3 - 2 = 12 - 2 = 10$. This matches the recursion.

For $n=3$, we have $a_3 = 3a_2 + 4 = 3(10) + 4 = 30 + 4 = 34$.

Using the formula, $a_3 = 4 \cdot 3^{3-1} - 2 = 4 \cdot 3^2 - 2 = 4 \cdot 9 - 2 = 36 - 2 = 34$. This matches the recursion.

3. Final Answer

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