Exercise · ZW-HOMOGENEOUS-LINEAR-RECURRENCE-008 · ★★☆ Intermediate

Solve \(a_n=5a_{n-1}-6a_{n-2}\) with \(a_0=1\) and \(a_1=5\).

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  • C. \(a_n=3^{n+1}-2^{n+1}\)

Explanation

With roots 2 and 3, \(a_n=\alpha2^n+\beta3^n\). From \(\alpha+\beta=1\) and \(2\alpha+3\beta=5\) we get \(\beta=3,\ \alpha=-2\). Check: \(a_2=5\cdot5-6=19=27-8\).

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Review the concept: Homogeneous linear recurrence

A homogeneous linear recurrence of order \(k\) expresses each term \(a_n\) as a combination of the previous \(k\) terms with constant coefficients \(c_1,\dots,c_k\), with no independent term.

\[a_n=c_1a_{n-1}+c_2a_{n-2}+\cdots+c_ka_{n-k},\quad c_k\neq0\]

More practice problems: homogeneous linear recurrence

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