Closed form solution of recurrence relation I am asked to solve following problem
Find a closed-form solution to the following recurrence: 
$\begin{align}
    x_0  &= 4,\\
    x_1  &= 23,\\

    x_n &= 11x_{n−1} − 30x_{n−2} \mbox{ for } n \geq  2.
\end{align}$
When I have searched what does mean  closed-form solution, wikipedia gives me answer  that it is expressed by the following  statement

An expression is said to be a closed-form expression if it can be expressed analytically in terms of a bounded number of certain "well-known" functions. 

So in this case we must find functions bounds or what? Please help me.
 A: Hint $ $ Let $\rm\:S\:$ be the shift-operator $\rm\:S\ f(n) = f(n\!+\!1).\:$ Then your recurrence factors as follows
$$\begin{align} 0\ &\rm =\ f(n\!+\!2) - 11\ f(n\!+\!1) + 30\ f(n)\\[.2em]
&\rm =\ (S^2 - 11\ S + 30)\,\ f(n)\\[.2em]
&\rm =\ (S-5)\,(S-6)\,\ f(n)\end{align}\qquad$$
Now $\rm\:(S-6)\ f(n) = 0\! \iff\! f(n\!+\!1) = 6\,f(n) \!\iff\! f(n) = c\,6^n\ $ for $\rm\,c\,$ constant, i.e. $\rm\:S\,c = c$.
and $\rm\: \ (S-5)\ f(n) = 0 \!\iff\! f(n\!+\!1) = 5\,f(n) \!\iff\! f(n) = d\,5^n\ $ for $\rm\,d\,$ constant.
Since $\rm\:S-5\:$ and $\rm\:S-6\:$ commute, we infer that $\rm\, c\, 6^n + d\, 5^n\,$ is a solution for all constants $\rm\,c,d.\,$ Now plug in the known initial conditions $\rm\:f(0) = x_0,\ f(1) = x_1\:$ to solve for the unknowns $\rm\:c,d\:.$
Simple linear algebra (using the Casoratian, a discrete analog of the Wronsian) shows that these  solutions span the solution space.
A: Use generating functions. Define $X(z) = \sum_{n \ge 0} x_n z^n$, rewrite your recurrence without subtractions in indices:
$$
x_{n + 2} = 11 x_{n + 1} - 30 x_n
$$
Multiply the recurrence by $z^n$, sum over all valid values of $n$ (i.e., $n \ge 0$), and recognize the resulting sums:
$$
\frac{X(z) - x_0 - x_1 z}{z^2} = 11 \frac{X(z) - x_0}{z} - 30 X(z)
$$
Solve for $X(z)$, write as partial fractions:
$$
X(z) = \frac{3}{1 - 6 z} + \frac{1}{1 - 5 z}
$$
Two geometric series, so:
$$
x_n = 3 \cdot 6^n + 5^n
$$
A: What you are being asked for is an equation that has $x_n$ on the left side and a formula in $n$ on the right side containing just familiar functions like polynomials and exponentials and only finitely many of them. Whoever asked you to solve the problem probably also provided a method for solving such problems, but here goes: 
First, let's write your problem in a better format: $x_0=4$, $x_1=23$, $x_n=11x_{n-1}-30x_{n-2}$. 
Now, suppose there is a solution of the form $x_n=c^n$ for some $c$. (Why do we make such a supposition? Because we've been here before, and we know it will work.) Then the equation says $c^n=11c^{n-1}-30c^{n-2}$, which simplifies to $c^2-11c+30=0$, which undoubtedly you are able to solve. You'll get two values of $c$ that work, let's call them $c_1$ and $c_2$, and then anything of the form $Ac_1^n+Bc_2^n$ will work also, for any numbers $A$ and $B$. If you're clever in choosing $A$ and $B$, you'll get $x_0=4$ and $x_1=23$.  
A: Please note this is my first time answering, so help me improve by keeping comments constructive
This is the characteristic polynomial method for finding a closed form expression of a recurrence relation, similar and dovetailing other answers:
given: $f(0)=v_1$, $f(1)=v_2$, ..., $f(d-1)=v_{d-1}$ and  $a_df(n) + a_d−1f(n − 1) + · · · + a_0f(n-d) = 0$ for all $n ≥ 0$
Note, you likely need to rewrite the recurrence relation, for example, as your recurrence relation presents
$x_n = 11x_{n-1} - 30x_{n-2}$ becomes $x_n-11x_{n-1}+30x_{n-2}=0$
The characteristic polynomial of this recurrence relation is of the form:
$q(x) = a_dx^d + a_{d−1}x^{d−1} + · · · + a_1x + a_0$
Now it's easy to write a characteristic polynomial using the coefficents $a_d$,$a_{d-1}$, ..., $a_0$:
$q(r)=r^2-11r+30$
Since $q(r) = 0$, the geometric progression $f(n) = r^n$ satisfies the implicit recurrence.
IF the roots of the characteristic equation are distinct, 
$f(n) = λ_1r_1^n + λ_2r_2^2 + · · · + λ_dr_d^n$, where $λ_1, . . . , λ_d$ are arbitrary complex numbers.
In this case, we have:
$q(r)=r^2-11r+30$ 
$q(r)=(r-5)(r-6)$ 
$r_1=5$ and $r_2=6$
So the general solution is:
$x_n=λ_15^n + λ_26^n$
Use the initial values to get two equations with two unknowns:
$f(0)=λ_15^0 + λ_26^0$ 
$4=λ_1 + λ_2$
$f(1)=λ_15^1 + λ_26^1$ 
$23=5λ_1 + 6λ_2$
Solving this, we have $λ_1 = 1$ and  $λ_2=3$, so the solution is
$x_n=5^n + 3*6^n$

IF the roots are NOT distinct, we need to think a bit further, but we'll leave that for another time.
A: Starting from the formula: $ x_n = 5^n + 3 \cdot 6^n $
Another way to see that same formula, is:
$$ x_n = 3^{1/2} \cdot 30^{n/2} \cdot (3^{1/2} \cdot (6/5)^{n/2}+3^{-1/2}(6/5)^{-n/2}) $$
Let: $$ t_n = \frac {ln(3) + ln(30) \cdot n} {2} $$
Let: $$ v_n = \frac {ln(3) + ln(6/5) \cdot n} {2} $$ 
Then:
$$ x_n = 2 \cdot exp(t_n) \cdot cosh(v_n) $$
Note: The numbers 6 and 5 of teh formula, is the two roots of: $$ z^2 -11z +30 = 0 $$
