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?

You have
$$a_n=a_n=a_{n-1}-6\cdot 3^{n-1}=a_{n-2}-6\cdot 3^{n-2}-6\cdot 3^{n-1}=\dotsm,$$
so you can prove with an easy induction that
$$a_n = a_0-6\sum_{k=0}^{n-1} 3^k=-6\frac{3^n-1}{3-1}.$$
