We are given a recurrence relation $a_n = 3a_{n-1} - 20$ for $n \ge 2$ and the initial condition $a_1 = 9$. We want to find a closed-form expression for $a_n$.

1. Problem Description

We are given a recurrence relation $a_n = 3a_{n-1} - 20$ for $n \ge 2$ and the initial condition $a_1 = 9$. We want to find a closed-form expression for $a_n$.

2. Solution Steps

First, we look for a particular solution of the form $a_n = c$, where $c$ is a constant. Substituting this into the recurrence relation, we get:

$c = 3c - 20$

$2c = 20$

$c = 10$

Now, we consider the homogeneous recurrence relation $a_n = 3a_{n-1}$. The general solution of this recurrence relation is $a_n = A(3)^{n-1}$, where $A$ is a constant.

Thus, the general solution to the non-homogeneous recurrence relation is $a_n = A(3)^{n-1} + 10$.

Using the initial condition $a_1 = 9$, we can find the value of $A$:

$a_1 = A(3)^{1-1} + 10 = 9$

$A(3)^0 + 10 = 9$

$A(1) + 10 = 9$

$A = -1$

Therefore, the closed-form expression for $a_n$ is $a_n = -1(3)^{n-1} + 10$.

3. Final Answer

$a_n = 10 - 3^{n-1}$