We need to find the general term $a_n$ for two sequences. a. $a_1 = 1$, $a_{n+1} = 4a_n + 9$ b. $a_1 = 1$, $a_2 = 1$, $a_{n+2} = a_{n+1} + 6a_n$

1. Problem Description

We need to find the general term $a_n$ for two sequences.

a. $a_1 = 1$, $a_{n+1} = 4a_n + 9$

b. $a_1 = 1$, $a_2 = 1$, $a_{n+2} = a_{n+1} + 6a_n$

2. Solution Steps

a. $a_1 = 1$, $a_{n+1} = 4a_n + 9$

This is a linear non-homogeneous recurrence relation of first order.

First, consider the homogeneous part: $a_{n+1} = 4a_n$.

The solution is of the form $a_n = A \cdot 4^n$, where $A$ is a constant.

Next, we look for a particular solution of the non-homogeneous equation.

Since the non-homogeneous part is a constant, we assume a constant solution, $a_n = C$.

Then $C = 4C + 9$, which gives $3C = -9$, so $C = -3$.

The general solution is $a_n = A \cdot 4^n - 3$.

We use the initial condition $a_1 = 1$ to find $A$:

$1 = A \cdot 4^1 - 3$, so $4A = 4$, which gives $A = 1$.

Therefore, $a_n = 4^n - 3$.

b. $a_1 = 1$, $a_2 = 1$, $a_{n+2} = a_{n+1} + 6a_n$

This is a linear homogeneous recurrence relation of second order with constant coefficients.

The characteristic equation is $r^2 - r - 6 = 0$.

Factoring, we get $(r-3)(r+2) = 0$, so the roots are $r_1 = 3$ and $r_2 = -2$.

The general solution is of the form $a_n = A \cdot 3^n + B \cdot (-2)^n$.

We use the initial conditions $a_1 = 1$ and $a_2 = 1$ to find $A$ and $B$:

$a_1 = 3A - 2B = 1$

$a_2 = 9A + 4B = 1$

Multiplying the first equation by 2, we get $6A - 4B = 2$.

Adding this to the second equation, we get $15A = 3$, so $A = \frac{1}{5}$.

Substituting $A = \frac{1}{5}$ into the first equation, we get $3(\frac{1}{5}) - 2B = 1$, so $\frac{3}{5} - 2B = 1$.

Thus, $2B = \frac{3}{5} - 1 = -\frac{2}{5}$, which gives $B = -\frac{1}{5}$.

Therefore, $a_n = \frac{1}{5} \cdot 3^n - \frac{1}{5} \cdot (-2)^n = \frac{3^n - (-2)^n}{5}$.

3. Final Answer

a. $a_n = 4^n - 3$

b. $a_n = \frac{3^n - (-2)^n}{5}$