Induction Proof - Number Sequence

Question

I tried this proof and my work was a bit different from how the solution solved it. I'd appreciate any feedback on my attempt:

Base case: n = 3, $a_3 ≤ 3^3$ = 25 ≤ 27, holds true

Induction Hypothesis: Assume S(k) holds for each number 3...k.

Proof of k+1:

$a_{k+1}$ = $2a_k$ + $a_{k-1}$

$< 2(3^k)+3^{k-1}$

$=3^{k+1}(\frac23+\frac19)$

$<3^{k+1}$

Answer 1

Your proof is correct.

You may enhance it to

$$a_{k+1} = 2a_k + a_{k−1} \le$$

$$2(3^k)+3^{k−1}=$$

$$=3^{k+1}(2/3+1/9) = (7/9)3^{k+1} <3^{k+1}$$

Answer 2

Your proof is not correct.

You have $n=3$ as your only base case, but you should have $n=4$ as another base case. Indeed, note that your step cannot establish $S(4)$, because it can only do that if $S(2)$ holds ... which it does not.