# Recursive sequence.square root 3. Find the recursion ratio [duplicate]

$$\begin{cases} {a_n} = 3a_{n-1} - 3a_{n-2} + a_{n-3}\\ {a_0} = 1, a_1=2, a_2=5 \end{cases}$$

Hey everyone.. So far I have solved only questions of 2 root kind. I tried to solve this one but no success.. I dont know what do to after i found out:

$$x^{3} - 3x^{2} + 3{x} - 1 = 0$$ equals to: $$(x-1)^{3}$$

Aaaand lost.

• Oct 6 '19 at 16:33
• @MartinR its different because here i need to solve the recursion ratio Oct 6 '19 at 16:37

• When root is simple you get $$(a)x^n$$
• When root is double you get $$(an+b)x^n$$
• When root is triple you get $$(an^2+bn+c)x^n$$

And so on...

Here $$x=1$$ is triple root, so the solution is $$a_n=(an^2+bn+c)\times 1^n=an^2+bn+c$$

Now search for initial conditions.

$$\begin{cases} a_0=c=1\\ a_1=a+b+c=2\\ a_2=4a+2b+c=5\end{cases}\iff \begin{cases} a=1\\ b=0\\ c=1\end{cases}\quad$$ and you get $$a_n=n^2+1$$

• THANK YOU! amazing Oct 6 '19 at 17:05

Here you can use $$b_n=a_n-a_{n-1}$$ so that $$b_n=2b_{n-1}-b_{n-2}$$ and you have reduced the problem to something you should be able to solve.

If you have a multiple root $$\alpha$$ of order $$m+1$$ (here the root $$x=1$$ of order $$3$$) you will find that the test solution $$a_n=p(n)\alpha^n$$ is the one to use, where $$p(n)$$ is a polynomial of degree $$m$$. Here you would try $$p(n)=An^2+Bn+C$$ (a quadratic). That should accord with what you know about double roots.

• Exactly. thanks for that. I will try with it now. one thing - if it was from root 4, then i should have irhgt $p(n)=An^{3}+Bn^{2}+Cn+D$ ? Oct 6 '19 at 16:39
• @LorinSherry If you have a quadruple root you have a cubic as you have suggested. Oct 6 '19 at 16:40
• alright. thank you Oct 6 '19 at 16:48