Finding an explicit formula for a recursive sequence. How to show that the recurrent formula 
$$A_n=A_{n-1} + A_{n-2} +4.$$
gives a sequence of the form      $f(n)=cr^n+cr^n$?
The only way we are allowed to solve it, is with the quadratic formula $(r^2-r-1)$...
 A: Hint: Consider $B_n = A_n+4$. Then $B_n=B_{n-1}+B_{n-2}$, which is a famous recurrence.
A: Consider $B_n=A_n+4$, to get $B_n=B_{n-1}+B_{n-2}$.
Now you can solve the chactaristic equation $r^2-r-1$.
This gives $r=\frac{1+\sqrt{5}}{2}$, $r=\frac{1-\sqrt{5}}{2}$.
Therefore we have $$B_n= A \left(\frac{1+\sqrt{5}}{2}\right)^n+ B \left(\frac{1-\sqrt{5}}{2}\right)^n$$
Therefore 
$$A_n= A \left(\frac{1+\sqrt{5}}{2}\right)^n+ B \left(\frac{1-\sqrt{5}}{2}\right)^n -4$$
Where $A,B$ depend on your initial conditions. 
A: More generally,
suppose that
$a_n
=\sum_{i=1}^k c_i a_{n-k} + u
$.
To get rid of the
constant term,
so this will be a
homogeneous recurrence,
let
$a_i
=b_i-v
$.
Then
$b_n-v
=\sum_{i=1}^k c_i (b_{n-k}-v) + u
=\sum_{i=1}^k c_i b_{n-k}-v\sum_{i=1}^kc_i + u
$
or
$b_n
=\sum_{i=1}^k c_i b_{n-k}-v(-1+\sum_{i=1}^kc_i) + u
$
Therefore,
if we choose
$v = \dfrac{u}{-1+\sum_{i=1}^kc_i }
$,
this is what we want.
Of course this assumes that
$\sum_{i=1}^kc_i 
\ne 1
$.
If
$\sum_{i=1}^kc_i 
= 1
$,
we have to do something else.
