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Suppose $h_0, h_1, h_2, \dots$ is a sequence defined:

$\qquad h_0=1, h_1=2, h_2=3$

$\qquad h_k=h_{k-1}+h_{k-2}+h_{k-3}, \forall k \in \mathbb{Z}\wedge k\ge 3$

Prove that $h_n \le 3^n,\forall n \in \mathbb{Z} \wedge n \ge 0$.


My basis is:

$(h_0 = 1) \le 3^0$ is true.

$(h_1 = 2) \le 3^1$ is true.

$(h_2 = 3) \le 3^2$ is true.

$(h_3 = 6) \le 3^3$ is true.

My inductive hypothesis is:

$h_k \le 3^k$ is true, for $3 \le k < n, n \in \mathbb{Z}$

$h_{k-1} + h_{k-2} + h_{k-3} \le 3^n$

Looking at the $k+1^{th}$ term, I have:

$h_{k+1} = h_{k} + h_{k-1} + h_{k-2}$

I'm really unsure where to go from here to prove the $k+1$ term is less than or equal to $3^n$. I can expand the $h_k$ term to get $2 * h_{k-1} + 2 * h_{k-2} + h_{k-3}$, but that doesn't seem to get me anywhere.

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    $\begingroup$ Hit: $3^n+3^n+3^n = 3^{n+1}$ $\endgroup$ – JavaMan Oct 12 '18 at 4:09
  • $\begingroup$ @JavaMan Oh. Would it make sense to compare $h_k \le 3^n$, $h_{k-1} \le 3^n$, $h_{k-2} \le 3^n$ and to say that it's true because of the inductive hypothesis? Or is that not valid to say? $\endgroup$ – Kevin Evans Oct 12 '18 at 4:26
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If $h_k\lt3^k$ then:

$$h_{k+1}=h_{k}+h_{k-1}+h_{k-2}<3^{k}+3^{k-1}+3^{k-2}<3^{k}+3^{k}+3^{k}=3^{k+1}$$

...or, in short:

$$ h_{k+1}<3^{k+1}$$

...which completes the induction step.

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