We are given a second-order linear homogeneous recurrence relation and asked to find a general formula for the $n$th term $a_n$. The recurrence relation is given by $a_{n+2} = a_{n+1} + 6a_n$, with initial conditions $a_1 = -1$ and $a_2 = 1$.

1. Problem Description

We are given a second-order linear homogeneous recurrence relation and asked to find a general formula for the $n$th term $a_n$. The recurrence relation is given by $a_{n+2} = a_{n+1} + 6a_n$, with initial conditions $a_1 = -1$ and $a_2 = 1$.

2. Solution Steps

First, we write down the characteristic equation of the given recurrence relation. The recurrence relation is $a_{n+2} = a_{n+1} + 6a_n$. We can rewrite this as $a_{n+2} - a_{n+1} - 6a_n = 0$.

The characteristic equation is obtained by replacing $a_{n+k}$ with $r^k$, where $r$ is a constant. Thus, we have:

$r^2 - r - 6 = 0$

Next, we solve for $r$. We can factor the quadratic equation as follows:

$(r - 3)(r + 2) = 0$

This gives us the roots $r_1 = 3$ and $r_2 = -2$.

Since the roots are distinct, the general solution of the recurrence relation is of the form:

$a_n = A(3)^n + B(-2)^n$, where A and B are constants.

Now, we use the initial conditions to find the values of A and B. We are given $a_1 = -1$ and $a_2 = 1$.

For $n=1$, we have $a_1 = A(3)^1 + B(-2)^1 = 3A - 2B = -1$.

For $n=2$, we have $a_2 = A(3)^2 + B(-2)^2 = 9A + 4B = 1$.

We now have a system of two linear equations in two variables:

$3A - 2B = -1$

$9A + 4B = 1$

Multiply the first equation by 2:

$6A - 4B = -2$

Add this to the second equation:

$(9A + 4B) + (6A - 4B) = 1 + (-2)$

$15A = -1$

$A = -\frac{1}{15}$

Substitute $A = -\frac{1}{15}$ into the first equation:

$3(-\frac{1}{15}) - 2B = -1$

$-\frac{1}{5} - 2B = -1$

$-2B = -1 + \frac{1}{5} = -\frac{4}{5}$

$B = \frac{2}{5}$

So the general solution is:

$a_n = -\frac{1}{15}(3)^n + \frac{2}{5}(-2)^n$

$a_n = -\frac{1}{15}3^n + \frac{6}{15}(-2)^n$

$a_n = \frac{-3^n + 6(-2)^n}{15}$

3. Final Answer

$a_n = \frac{-3^n + 6(-2)^n}{15}$