Solve recursion $a_n=a_{n-1}-6\cdot3^{n-1}$ for $n>0, a_0=0$ $a_n=a_{n-1}-6\cdot3^{n-1}$ for  $n>0,  a_0=0$
So I calculate first terms
$a_0=0$
$a_1=-6$
$a_2=-24$
$a_3=-78$
I don't see any relation so 
$a_n=a_{n-1}-6\cdot3^{n-1}$
$a_{n-1}=a_{n-2}-6\cdot 3^{n-2}$
. . .
$a_2=a_1-6\cdot3^{1}$
$a_1=a_0-6\cdot 3^{0}$
Not sure what to do next, Wolfram solves it in this way:
$a_n=-3\cdot(3^{n}-1)$
How do I get to this point?

Note that for all $n$ we have $$a_{n+1}-a_n = -6\cdot 3^n $$
so we have also: $$a_n-a_{n-1}=-6\cdot3^{n-1}$$
thus $$a_{n+1}-a_n = 3(-6\cdot3^{n-1}) = 3(a_n-a_{n-1})$$
{or divide this two equations:  $${a_{n+1}-a_n \over a_n-a_{n-1}}= {-6\cdot 3^n\over -6\cdot3^{n-1}} = 3$$}  
so you have to solve linear recurrence:
$$ a_{n+1}-4a_n+3a_{n-1}=0$$
CAn you do that?
