Answer
\[
\boxed{t_k = 3^k, \quad \text{for all integers }k \ge 0.}
\]
Work Step by Step
We have the recurrence
\[
t_k \;=\; 6\,t_{k-1} \;-\; 9\,t_{k-2},
\quad \text{for }k\ge 2,
\quad \text{with }t_0=1,\; t_1=3.
\]
---
## 1. Solve the Characteristic Equation
First, look for solutions of the form \(t^k\). Substituting \(t_k = r^k\) into the recurrence:
\[
r^k
\;=\; 6\,r^{k-1} \;-\; 9\,r^{k-2}.
\]
Divide by \(r^{k-2}\) (assuming \(r \neq 0\)):
\[
r^2
= 6\,r \;-\; 9,
\]
which rearranges to
\[
r^2 \;-\; 6\,r \;+\; 9 = 0.
\]
Recognize it as a perfect square:
\[
(r - 3)^2 = 0
\quad\Longrightarrow\quad
r = 3 \quad(\text{a repeated root}).
\]
---
## 2. General Form for a Double Root
When the characteristic equation has a repeated root \(r\), the general solution to the homogeneous recurrence is:
\[
t_k
= A\,r^k + B\,k\,r^k.
\]
Here, \(r=3\). So
\[
t_k
= A\,3^k + B\,k\,3^k.
\]
---
## 3. Use the Initial Conditions
We have \(t_0=1\) and \(t_1=3\).
1. **For \(k=0\)**:
\[
t_0
= A\,3^0 + B\,(0)\,3^0
= A
= 1.
\]
So \(A=1.\)
2. **For \(k=1\)**:
\[
t_1
= A\,3^1 + B\,(1)\,3^1
= 3A + 3B
= 3\,(A + B).
\]
But \(t_1=3\). Hence
\[
3(A + B) = 3
\quad\Longrightarrow\quad
A + B = 1.
\]
Since we already found \(A=1\), it follows that
\[
1 + B = 1
\quad\Longrightarrow\quad
B = 0.
\]
