# Can we express $p_n$ in terms of $p_0, p_1$ and $n$?

$p_0=a$, $p_1=b$, $bp_n=p_{n+1}+p_{n-1}$ express $p_n$ in terms of $a,b,n$.

Any help would be appreciated, because you guys are splendid.

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It's easier to see if you rewrite it as $p_{n+1} = bp_n - p_{n-1}$. Calculate the first few values, and you'll see a pattern. Prove it by induction. – Yuri Sulyma Oct 21 '11 at 2:01

You have a linear three-term recurrence. Rearrange as

$$p_{n+1}-bp_n+p_{n-1}=0$$

and you obtain the characteristic polynomial $x^2-bx+1=0$. This means that $p_n$ can be expressed as

$$p_n=cx_1^n+dx_2^n$$

where $x_1,x_2$ are the roots of the quadratic. You can determine $c$ and $d$ in terms of $a$ and $b$ by using the initial conditions and solving the resulting linear equations...

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how could you finish the problem if a=2, and b=3 ? – Victor Oct 21 '11 at 2:23
Victor, I think J. M. has told you what to do. Solve the quadratic to get $x_1,x_2$; find $c,d$ by using the formula J. M. gave you for $p_n$ and the initial conditions, that is, the values given for $p_0,p_1$. Have you tried that? Where did you get stuck? – Gerry Myerson Oct 21 '11 at 5:22
+1 to @J.M. for not finishing the problem. – Did Oct 21 '11 at 5:31
@Didier: Given OP's history, I consciously chose not to... – J. M. Oct 21 '11 at 10:07