Define a sequence $a_1, a_2, a_3, \ldots$ as follows: $a_1 = 1,$ $a_2 = 3,$ $a_k = a_{k-1} + a_{k-2},$ $\forall k \in \mathbb{Z}, k \geq 3$. This sequence is known as the Lucas sequence. Use strong mathematical induction to prove that $a_n \leq (\frac{7}{4})^n$ $\space \space \forall n \in \mathbb{Z}, n \geq 1$
We've started learning induction/strong induction, however, I'm still a little bit confused if I've successfully proven the above sequence via strong induction. What I have so far:
Let $P(n)$ be the statement that $a_n \leq (\frac{7}{4})^n, \forall n \in \mathbb{Z}, n \geq 1$. For the basis, we are given:
$a_1 = 1$
$a_2 = 3$
$a_k = a_{k-1} + a_{k-2}, \space \forall k \in \mathbb{Z}, k\geq 3$
Let $k$ be any integer $\geq 3$ and suppose $P(i)$ is the statement that $a_i \leq (\frac{7}{4})^i, \forall i \in \mathbb{Z}, 1 \leq i \leq k,$ then $P(i) \Longrightarrow P(k+1)$.
Do I simply just plug $i = k+1$ in to $a_i$ and $(\frac{7}{4})^i$ then compare like so?
$a_{k+1} = a_k + a_{k-1}$
By the inductive hypothesis:
$a_k + a_{k-1} \leq (\frac{7}{4})^k + (\frac{7}{4})^{k-1}$
$=(\frac{7^{k}}{4^{k}}) + (\frac{7^{k-1}}{4^{k-1}})$
$=(\frac{7^{k+1}}{4^{k+1}})(\frac{4}{7}) + (\frac{7^{k+1}}{4^{k+11}})(\frac{4^2}{7^2})$ by the quotient rule for exponents
$=(\frac{7^{k+1}}{4^{k+1}})[\frac{4}{7} + \frac{4^2}{7^2}]$ by factoring
$=(\frac{7}{4})^{k+1}[\frac{44}{49}] \leq (\frac{7}{4})^{k+1}$ as $\frac{44}{49}$ is clearly $ < 1$
