Let $_0, _1, _2, …$ be the sequence defined by the following recurrence relation: Let $_0, _1, _2, …$ be the sequence defined by the following recurrence
relation:
$_0 = 2$
$_1 = 2$
$_2 = 6$
$_ = 3_{−3}$ for $ ≥ 3$
Prove that  is even for any nonnegative integer .
a. The base cases are  = ,  = , and  = .
$_ = , _ = ,$ and $_ = $ are even.
Thus, the statement is true for  = ,  = , and  = .
b. Assume that $_$ is even for  ≤  ≤  and  ≥ .
That is, $_ = _$ for some integer .
c. Show that if the inductive hypothesis is true, then $_{+}$ is even.
d. $_{+} = _{−}$
$_{+} = ()$ (inductive hypothesis) -> From where does this come from
$_{+} = ()$
Let  be an integer such that  = .
Then, $_{+} = .$
Thus, $_{+}$ is even.
Therefore, by strong induction, the statement is true for any nonnegative
integer .
Can anyone please explain me how does $_ = _$ and also the proof part. From where does the inductive hypothesis come from in the proof. 

Even number is divisible by 2, so it is of the form $2i$ (not $2_i$) for some natural (or integer) $i$.
Probably easier to understand will be if you temporary forget induction. Because $a_0$, $a_1$, $a_2$ are even, their multiplications by 3 are also even, and their multiplication by 3 are also even, and so on. But it is an informal induction, which can be formalized as in your proof.
