How to find the general formula for this recursive problem? I want to ask about recursive problem.
Given: $$a_0= 11, a_1= -13,$$ and $$a_n= -a_{n-1} +2a_{n-2}.$$
What is the general formula for $$a_n$$ ?
I've already tried to find the first terms of this series. From there, I got:
$$a_2 = 35, a_3= -61,$$ and $$a_4= 131.$$
From there, I think I need to use the rule from arithmetic and geometric series to find the general formula that I want to find.
But, I cannot find the certain pattern from this series, because the differences is always changing, such that -24, 48, -96, 192. 
From there I think the general formula for a_n should be including (-1)ⁿ. But, how can we deal with series 24,48,96,192?
It seems the series is geometric, but how can we find the formula?
Thanks

Let me try. Putting $b_n = a_{n+1} + 2a_n$, for $n \geq 1$, one has $$b_0=9,$$ and $$b_{n-1} = b_{n-2},\ \mbox{ for } n \geq 2.$$
Then one has $b_n = b_0 = 9$, for all $n$.
So, $a_{n+1} + 2a_n = 9,$ for all $n$.
Putting $c_n = a_n - 3$, one has $c_{n+1} + 2c_n = 0$ and $c_0 = 8$.
Thus, $c_n = -2c_{n-1} = (-2)^n c_0 = 8.2^n.(-1)^n$.
We can get $a_n = 8.2^n.(-1)^n + 3$.
EDIT:
The way chooses $b_n$ is a trick. We can solve this problem as follows.
Consider the equation: $a^2+a-2=0$. It has two solutions: $a_1=1$ and $a_2=-2$. Then, $x_n = \alpha a_1^n + \beta a_2^n$. Indeed, one has
$$x_{n+2} + x_{n+1} - 2x_n = \alpha a_1^n(a_1^2+a_1-2) + \beta a_2^n(a_2^2+a_2-2) = 0.$$
Now, we find $\alpha$ and $\beta$.
We have $x_0 = \alpha + \beta = 11$, $x_1 = \alpha - 2\beta = -13$. Then, $\alpha = 3$ and $\beta = 8$. We get $$x_n = 3 + 8(-2)^n$$
