Recursively Defined Sequence/Examples/Term is Term of Index less 1 plus 6 times Term of Index less 2

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Theorem

Consider the integer sequence $\left\langle{a_n}\right\rangle$ defined recursively as:

$a_n = \begin{cases} 0 & : n = 0 \\ 1 & : n = 1 \\ a_{n - 1} + 6 a_{n - 2} & : \text{otherwise} \end{cases}$


$a_n$ has a closed-form expression:

$a_n = \dfrac {3^n - \left({-2}\right)^n} 5$


Proof

The proof proceeds by strong induction.

For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:

$a_n = \dfrac {3^n - \left({-2}\right)^n} 5$


Basis for the Induction

$P \left({0}\right)$ is the case:

\(\ds \dfrac {3^0 - \left({-2}\right)^0} 5\) \(=\) \(\ds \dfrac {1 - 1} 5\)
\(\ds \) \(=\) \(\ds 0\)
\(\ds \) \(=\) \(\ds a_0\)

Thus $P \left({0}\right)$ is seen to hold.


$P \left({1}\right)$ is the case:

\(\ds \dfrac {3^1 - \left({-2}\right)^1} 5\) \(=\) \(\ds \dfrac {3 - \left({-2}\right)} 5\)
\(\ds \) \(=\) \(\ds \dfrac {3 + 2} 5\)
\(\ds \) \(=\) \(\ds 1\)
\(\ds \) \(=\) \(\ds a_1\)

Thus $P \left({1}\right)$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $P \left({j}\right)$ is true, for all $j$ such that $0 \le j \le k$, then it logically follows that $P \left({k + 1}\right)$ is true.


This is the induction hypothesis:

$a_k = \dfrac {3^k - \left({-2}\right)^k} 5$

and:

$a_{k - 1} = \dfrac {3^{k - 1} - \left({-2}\right)^{k - 1} } 5$


from which it is to be shown that:

$a_{k + 1} = \dfrac {3^{k + 1} - \left({-2}\right)^{k + 1} } 5$


Induction Step

This is the induction step:

\(\ds a_{k + 1}\) \(=\) \(\ds a_k + 6 a_{k - 1}\) Definition of $a_n$
\(\ds \) \(=\) \(\ds \dfrac {3^k - \left({-2}\right)^k} 5 + 6 \times \dfrac {3^{k - 1} - \left({-2}\right)^{k - 1} } 5\)
\(\ds \) \(=\) \(\ds \dfrac {3^k - \left({-2}\right)^k + 6 \times 3^{k - 1} - 6 \left({-2}\right)^{k - 1} } 5\)
\(\ds \) \(=\) \(\ds \dfrac {3^{k - 1} \left({3 + 6}\right) - \left({-2}\right)^{k - 1} \left({-2 + 6}\right)} 5\)
\(\ds \) \(=\) \(\ds \dfrac {3^{k - 1} \times 9 - \left({-2}\right)^{k - 1} \times 4} 5\)
\(\ds \) \(=\) \(\ds \dfrac {3^{k - 1} \times 3^2 - \left({-2}\right)^{k - 1} \times \left({-2}\right)^2} 5\)
\(\ds \) \(=\) \(\ds \dfrac {3^{k + 1} - \left({-2}\right)^{k + 1} } 5\)

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Second Principle of Mathematical Induction.


Therefore:

$\forall n \in \Z_{\ge 0}: a_n = \dfrac {3^n - \left({-2}\right)^n} 5$

$\blacksquare$


Although this article appears correct, it's inelegant. There has to be a better way of doing it.
In particular: Given that this result was given as a question in a chapter on the Fibonacci numbers, one presumes there is a way to solve this by using Fibonacci numbers, or a Fibonacci-like approach. Maybe use generating functions.
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Sources

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