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

## Contents

## 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:

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

\(\displaystyle \) | \(=\) | \(\displaystyle 0\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a_0\) |

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

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

\(\displaystyle \dfrac {3^1 - \left({-2}\right)^1} 5\) | \(=\) | \(\displaystyle \dfrac {3 - \left({-2}\right)} 5\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac {3 + 2} 5\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a_1\) |

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:

\(\displaystyle a_{k + 1}\) | \(=\) | \(\displaystyle a_k + 6 a_{k - 1}\) | Definition of $a_n$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac {3^k - \left({-2}\right)^k} 5 + 6 \times \dfrac {3^{k - 1} - \left({-2}\right)^{k - 1} } 5\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac {3^k - \left({-2}\right)^k + 6 \times 3^{k - 1} - 6 \left({-2}\right)^{k - 1} } 5\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac {3^{k - 1} \left({3 + 6}\right) - \left({-2}\right)^{k - 1} \left({-2 + 6}\right)} 5\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac {3^{k - 1} \times 9 - \left({-2}\right)^{k - 1} \times 4} 5\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac {3^{k - 1} \times 3^2 - \left({-2}\right)^{k - 1} \times \left({-2}\right)^2} 5\) | |||||||||||

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

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

Therefore:

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

$\blacksquare$

## Sources

- 1997: Donald E. Knuth:
*The Art of Computer Programming: Volume 1: Fundamental Algorithms*(3rd ed.) ... (previous) ... (next): $\S 1.2.8$: Fibonacci Numbers: Exercise $39$