# 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?

• Welcome to TeX SX! Do you mean $a_n$ or $an$? – Bernard Jan 27 '19 at 13:07
• Hello, yes I meant first option, didn't know how to do it in tex, sorry. – Gorosso Jan 27 '19 at 13:14
• it's just a_{n}, a_{n-1}, &c. If there's only 1 character in subscript, the braces are not necessary. – Bernard Jan 27 '19 at 13:21

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

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.

This is just a Geometric Series:

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

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?

• ok so: $x^{2}-4x+3$ $\Delta =4$ $x1=1$ $x2=3$ Now I use following formula $a_n=ar^{n}+br^{n}$ $a_0=0=a+b$ $a_1=-6=a+3b$ $a=-b$ $-6=2b$ $b=-3$ $a=3$ $a_n=3*1^{n}-3*3^{n}$ – Gorosso Jan 27 '19 at 14:40
• Yes, that is correct! – Aqua Jan 27 '19 at 14:40
• Is that better? – Aqua Jan 27 '19 at 14:52
• What about now? – Aqua Jan 27 '19 at 15:51
• Just put first and last expression on one side, so you get 0 on other side – Aqua Jan 27 '19 at 16:54