how to work out a closed form of a sequence 
Consider the following linear recurrence sequence.

$x_1 = 11$,  $x_{n+1} = -0.8x_n + 9,\quad n = 1,2,3, \ldots.$
Find a closed form for this sequence.

To get an idea of what the closed form might look like, let's iterate the relation a few times:
$$
\begin{align}
x_n
&=9-.8x_{n-1}\\
&=9-.8(9-.8x_{n-2})\\
&=9-.8\cdot9+.8^2x_{n-2}\\
&=9-.8\cdot9+.8^2(9-.8x_{n-3})\\
&=9-.8\cdot9+.8^2\cdot9-.8^3x_{n-3}\tag{1}
\end{align}
$$
Looking at $(1)$, it appears that
$$
x_n=9\frac{1-(-.8)^k}{1+.8}+(-.8)^kx_{n-k}\tag{2}
$$
Equation $(2)$ can be verified using induction.

Verification: The case $k=1$ is just the given recursion: $x_n=9-.8x_{n-1}$.

Suppose $(2)$ is true for some $k$, then
$$
\begin{align}
x_n
&=9\frac{1-(-.8)^k}{1+.8}+(-.8)^kx_{n-k}\\
&=9\frac{1-(-.8)^k}{1+.8}+(-.8)^k(9-.8x_{n-k-1})\\
&=9\frac{1-(-.8)^k}{1+.8}+9(-.8)^k+(-.8)^{k+1}x_{n-k-1}\\
&=9\frac{1-(-.8)^{k+1}}{1+.8}+(-.8)^{k+1}x_{n-k-1}\\
\end{align}
$$
So $(2)$ is true for $k+1$.

Using $x_1=11$ and $k=n-1$, $(2)$ becomes
$$
\begin{align}
x_n
&=5(1-(-.8)^{n-1})+(-.8)^{n-1}11\\
&=5+6(-.8)^{n-1}\tag{3}
\end{align}
$$
