We are given a sequence $\{a_n\}$ with the first term $a_1 = 2$ and a recursive relation $a_{n+1} = 3a_n + 4$. We need to find a general formula for $a_n$ in terms of $n$.

1. Problem Description

We are given a sequence $\{a_n\}$ with the first term $a_1 = 2$ and a recursive relation $a_{n+1} = 3a_n + 4$. We need to find a general formula for $a_n$ in terms of $n$.

2. Solution Steps

The recursive formula is $a_{n+1} = 3a_n + 4$.

Let's assume $a_n = A \cdot 3^n + B$ for some constants $A$ and $B$.

Substituting this into the recursive relation, we have:

$A \cdot 3^{n+1} + B = 3(A \cdot 3^n + B) + 4$

$A \cdot 3^{n+1} + B = A \cdot 3^{n+1} + 3B + 4$

$B = 3B + 4$

$-2B = 4$

$B = -2$

So, $a_n = A \cdot 3^n - 2$.

Now, we use the initial condition $a_1 = 2$ to find $A$:

$a_1 = A \cdot 3^1 - 2 = 2$

$3A - 2 = 2$

$3A = 4$

$A = \frac{4}{3}$

Therefore, the general term is $a_n = \frac{4}{3} \cdot 3^n - 2$. We can rewrite this as $a_n = 4 \cdot 3^{n-1} - 2$.

3. Final Answer

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