Let us assume that $k$, $f(1)$, and $f(0)$ are constants. I am trying to find a closed form expression for $f(a)$, $a >= 2$ in terms of $k$, $f(1)$, and $f(0)$. I have to tried to recursively substitute values for $f(a)$ to try and generate a pattern. \begin{eqnarray} f(2) &=& k f(1) - f(0) \newline f(3) &=& (k^{2}-1) f(1) - k f(0) \newline f(4) &=& k(k^{2}-3) f(1) - (k^{2}-1) f(0) \newline f(5) &=& (k^{4}-3k^{2}+1) f(1) - k(k^{2}-2) f(0) \end{eqnarray} However, this is perhaps not the right way to approach this problem, and I'm looking for alternative ways to solve it?
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2it will be $f(n)=Ca^n+Db^n$ where $a$ and $b$ are roots of $x^2-kx+1=0$; cf. this – J. W. Tanner Feb 24 '20 at 18:51
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Here is an example you could use – rtybase Feb 24 '20 at 19:17
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1@J.W.Tanner I solved the recurrence relation using your approach. Thanks! – V-Red Feb 24 '20 at 19:22