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here are the steps I have done to try and find the general solution of this relation: $$ b_n = 3b_{n-1} - b_{n-3}\\ = b^n = 3b^{n-1} - b^{n-3}$$ then divide by $b^{n-3}$ to get $$b^3 = 3b^2 - 1$$ then bring to other side $$b^3 - 3b^2 + 1 = 0$$ This is our characteristic equation. However, now I am stuck because I cannot solve for $b$, which will be needed for the general solution. If anyone can help me solve this problem help it would be greatly appreciated For a recurrence relation $a_n$ the general solution should be $$ a_n = A_1a^n_1 + A_2a^n_2 + ... + A_ra^n_r $$ where $r$ is the number of roots for our characteristic equation.

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  • $\begingroup$ What Will means is that it is unclear what you mean by "solve the recurrence." All those question marks are rude and smack of mockery, @WillJagy If you must post a one-word comment, there are other ways to do so which are not rude. I sometimes use ${}{}{}{}$ $\endgroup$
    – Thomas Andrews
    Commented May 4 at 1:08
  • $\begingroup$ The question replaces subscripts by superscripts (powers) in the beginning of the calculation. Recurrence equations do not work that way. $\endgroup$
    – Steen82
    Commented May 4 at 1:13
  • $\begingroup$ i have made some edits to clarify the question. I am trying to get to the general solution. Also because this is a linear recurrence relation problem we can substitute a_n = a^n for each in the relation. $\endgroup$
    – sor3n
    Commented May 4 at 1:30
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    $\begingroup$ This is probably the best you can do with this method: The general solution is $b_n = A_1a_1^n+A_2 a_2^n+A_3 a_3^n$, where $a_1,a_2,a_3$ are the three (real) zeros of $b^3-3b^2+1=0$. $\endgroup$
    – GEdgar
    Commented May 4 at 1:32
  • $\begingroup$ And there are formulas for the zeros of a cubic, much like there is the quadratic formula for zeros of a quadratic, only the cubic is considerably more complicated. Said formulas are probably available on this here website, and certainly available in dozens of textbooks and on hundreds of other websites. $\endgroup$
    – Gerry Myerson
    Commented May 4 at 1:40

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