<s>[INST] Use complete induction to prove that $a_n < 2^n$ for every integer $n \geq 2$ Define the sequence of integers $a_0, a_1, a_2, \cdots$ as follows
$$ a_i =
\begin{cases} 
      i+1 & 0 \leq i \leq 2 \\
      a_{i-1} + a_{i-2} + 2a_{i-3} & i > 2 \\
   \end{cases}
$$
Use complete induction to prove that $a_n < 2^n$ for every integer $n \geq 2$
I will prove $a_n < 2^n, \forall n \geq 2$ by using complete induction
Base Case: Three cases $n = 2, 3, 4$
let $n = 2$
$a_{n} = a_{2} = 2 + 1 = 3 < 4 = 2^2 = 2^n$ By definition, and holds
let $n = 3$
$a_{n} = a_{3} = 3 + 2 + 1 = 6 < 8 = 2^3 = 2^n$ By definition, and holds
let $n = 4$
$a_{n} = a_{4} = 4 + 3 + 2 = 9 < 16 = 2^4 = 2^n$ By definition, and holds
Inductive step: let $n > 4$. Suppose $a_j < 2^j$ whenever $2 \leq j < n$. [Inductive hypothesis]
What to prove: $a_n < 2^n$
$a_{n} = a_{i-1} + a_{i-2} + 2a_{i-3}$ [By definition]
$< 2^{n-1} + 2^{n-2} + 2^{n-3+1}$ [By Inductive hypothesis]
$= 2^{n-1} + 2^{n-2} + 2^{n-2}$
$= 2^{n-1} + 2^{n-2 + 1}$
$= 2^{n-1 + 1}$
$= 2^n$
as wanted.
Would this be correct?
 [/INST] The second part is fine. But the base case is not just the $n=2$ case. You should have checked it for $n=0$, and $n=1$ too. 
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