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

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.

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  • 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 $a_n=n^2+1$

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  • $\begingroup$ THANK YOU! amazing $\endgroup$ Oct 6 '19 at 17:05
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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.

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

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