The problem asks us to find the general term $a_n$ of the given 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

The problem asks us to find the general term $a_n$ of the given 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. Let $a_n = b_n + c$. Then

$b_{n+1} + c = 4(b_n + c) + 9 = 4b_n + 4c + 9$

$b_{n+1} = 4b_n + 3c + 9$

Let $3c+9=0$, so $c=-3$.

Then $b_{n+1} = 4b_n$, which means $b_n = b_1 4^{n-1}$.

Since $a_1 = b_1 + c = 1$, $b_1 = 1 - c = 1 - (-3) = 4$.

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

Then $a_n = b_n + c = 4^n - 3$.

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

This is a homogeneous linear recurrence relation with constant coefficients.

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

$(r-3)(r+2) = 0$, so $r_1 = 3$, $r_2 = -2$.

Therefore, $a_n = A(3)^n + B(-2)^n$ for some constants A and B.

We use $a_1 = 1$ and $a_2 = 1$ to solve for A and B.

$a_1 = 3A - 2B = 1$

$a_2 = 9A + 4B = 1$

Multiply the first equation by 2: $6A - 4B = 2$.

Add this to the second equation: $15A = 3$, so $A = \frac{1}{5}$.

Then $3(\frac{1}{5}) - 2B = 1$, so $\frac{3}{5} - 2B = 1$, $2B = \frac{3}{5} - 1 = -\frac{2}{5}$.

So $B = -\frac{1}{5}$.

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

3. Final Answer

(a) $a_n = 4^n - 3$

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