Consider the sequence $\{a_n\}$, where $a_0 = 1$; $a_1 = 2$; $a_2 = 3$ and $a_n = a_{n-1} + a_{n-2} + a_{n-3}$; $n$ belongs to set $\mathbb{Z}^+$; where $n \ge 3$.
(b) Prove by mathematical induction that for all $n \in \mathbb{N}$ , we have that $a_n \le 3^n$.
In this case all you need from the inductive hypothesis is (I) $a_{n-1}\leq 3^{n-1}, a_{n-2}\leq 3^{n-1}$, and $a_{n-3}\leq 3^{n-1}$. Notice that this is a weaker statement than $a_{n-i}\leq 3^{n-i}$ for $i\in\{1,2,3\}$, which is still a weaker statement than $a_{k}\leq 3^{k}$ for all $k<n$. This last statement is the inductive hypothesis if we are using strong induction (or complete induction).
Let's try and prove this with statement (I). Notice for this inductive hypothesis we need to prove our base case for not just $a_{0}$, but also for $a_{1}$ and $a_{2}$. This is easy: $a_{0}<a_{1}<a_{2}\leq 3^{2}$. Now consider the following statement:
(II) If $a_{n-1}\leq 3^{n-1}, a_{n-2}\leq 3^{n-1}$ and $a_{n-3}\leq 3^{n-1}$ then $a_{n}\leq 3^{n}$.
If we can prove (II) then notice how our base case and (2) imply that $a_{3}\leq 3^{3}$. Additionally, because we have a strictly increasing sequence, this would also imply that $a_{1},a_{2},a_{3}\leq 3^{3}$ so then (II) would imply that $a_{4}\leq 3^{4}, a_{5}\leq 3^{5},\ldots,a_{n}\leq3^{n},\ldots$
Proving (II) should be a straightforward application of how the sequence $\{a_{n}\}$ is defined. You could choose to use strong induction as mentioned above, but your base case will still need to be $a_{0}\leq1, a_{1}\leq 3, a_{2}\leq 9$. Or, as mentioned in an above comment, becase $\{a_{n}\}$ is strictly increasing, it suffices to show that $a_{2}\leq 9$ and we can use weak induction.
