Solve recursion $a_n=a_{n-1}-6\cdot3^{n-1}$ for $n>0, a_0=0$

Question

$a_n=a_{n-1}-6\cdot3^{n-1}$ 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?

Answer 1

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?

Answer 2

This is just a Geometric Series:

$$a_n=-6\sum_{i=1}^n3^{i-1}=-6\sum_{i=0}^{n-1}3^i=-6\times \frac {3^n-1}{3-1}=-3\times (3^n-1)$$

Answer 3

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 $$a_n = a_0-6\sum_{k=0}^{n-1} 3^k=-6\frac{3^n-1}{3-1}.$$

Answer 4

hint...consider $$\sum_{r=0}^{r=n}(a_{r+1}-a_r)=\sum_{r=0}^{r=n}-6\cdot3^r$$

The LHS is a telescoping series and the RHS is a geometric series.