My task: $a_n=a_{n-1}+6a_{n-2}+30$ for $n\geq2$, $a_0=0$, $a_1=-10$

My solution $x^{2}-x-6$


$x1=-2 $


So I am gonna use following formula: $a_n=ar^{n}+br^{n}$







I calculate $a_2$


Which is correct with above formula


but for $a_3$ and above results are not matching, am I doing something wrong here?


The $+30$ is the problem. Find $c$ such that $b_n:=a_n+c\implies b_n=b_{n-1}+6b_{n-2}$, then use your usual techniques.

Edit to add detail:

$$b_n=b_{n-1}-c+6(b_{n-2}-c)+30+c=b_{n-1}+6b_{n-2}+30-6c$$ Set $c=5$ so $$b_n=b_{n-1}+6b_{n-2}\implies\exists a,\,b:\,b_n=a(-2)^n+b3^n.$$From $b_0=5,\,b_1=-5$, you can find $a,\,b$. Then $a_n=b_n-5$.

  • $\begingroup$ I don't really get it, could you show me how its supposed to look in my case? That's the best way for me to understand it. $\endgroup$ – Gorosso Jan 27 '19 at 16:41
  • $\begingroup$ @Gorosso See my edit. $\endgroup$ – J.G. Jan 27 '19 at 17:15
  • $\begingroup$ Where does c=5 come from? Also what I am supposed to do with $a_n=b_n-5$? $\endgroup$ – Gorosso Jan 27 '19 at 18:00
  • $\begingroup$ @Gorosso The technique you used originally only works when there isn't a constant in the recursion relation, so I chose $c$ to delete it. Once you have a formula for $b_n$, you have one for $a_n$. $\endgroup$ – J.G. Jan 27 '19 at 18:14
  • $\begingroup$ Now results are matching, so I guess I did everything according to your tips. But I am having some issue in similar reccurence using your method, could you check it out? math.stackexchange.com/questions/3090006/… $\endgroup$ – Gorosso Jan 27 '19 at 19:22

Since for all $n$ we have $$a_n-a_{n-1}-6a_{n-2}=30$$ we have also


so $$a_n-a_{n-1}-6a_{n-2}=a_{n-1}-a_{n-2}-6a_{n-3}$$

and thus we get l.r. :


and so on...


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