Closed form for $t_n = 6t_{n-1}-9t_{n-2}$ where $t_0 = 5$ and $t_1 = 9$

Question

Consider the sequence defined by $$ \begin{cases} t_0=5\\ t_1=9\\ t_n=6t_{n-1}-9t_{n-2} & \text{if }n\ge 2 \end{cases} .$$ Find a closed form for $t_n$.

Your response should be a formula in terms of $n$, and should not contain terms such as $t_n,$ $t_{n-1},$ and so on. Do not include $``t_n=\text{''}$ in your response.

I tried forming a sequence by taking some values for $n$ and finding $t_n$. Once that was done, I moved on to find some pattern between $n$ and $t_n$ but couldn't find any. Here's the sequence:

$$\begin{array}{c || c *5{| c}} n & 0 & 1 & 2 & 3 & 4 \\ \hline t_n & 5 & 9 & 9 & -27 & -243 \end{array}$$

What am I suppose to do?

Answer 1

Hint. Make the ansatz $$t_n=q^n$$ and for your work: the solution is given by $$t_n=3^n(5-2n)$$

Answer 2

Note that the characteristic polynomial of your homogeneous linear recurrence is $z^2-6z+9=(z-3)^2$. Therefore the solution has the form $$t_n=(An+B)3^n$$ where $A$ and $B$ are constants to be found.

Answer 3

Another widely used approach is using generating functions, i.e. $$f(x)=\sum\limits_{n=0}\color{red}{t_n}x^n= 5+9x+\sum\limits_{n=2}t_nx^n= 5+9x+\sum\limits_{n=2}\left(6t_{n-1}-9t_{n-2}\right)x^n=\\ 5+9x+6x\left(\sum\limits_{n=2}t_{n-1}x^{n-1}\right)-9x^2\left(\sum\limits_{n=2}t_{n-2}x^{n-2}\right)=\\ 5+9x+6x\left(\sum\limits_{n=1}t_{n}x^{n}\right)-9x^2\left(\sum\limits_{n=0}t_{n}x^{n}\right)= 5+9x+6x\left(f(x)-5\right)-9x^2f(x)$$ or $$f(x)=5+9x+6x\left(f(x)-5\right)-9x^2f(x) \iff \\ f(x)=\frac{5-21x}{1-6x+9x^2}= \frac{5-21x}{(1-3x)^2}= \frac{7}{1-3x}-\frac{2}{(1-3x)^2}=\\ 7\left(\sum\limits_{n=0}3^nx^n\right)-2\left(\sum\limits_{n=0}(n+1)3^nx^n\right)=\\ \sum\limits_{n=0}\color{red}{\left(7-2(n+1)\right)3^n}x^n$$ and $$t_n=\left(5-2n\right)3^n$$