How prove this sequence $a_{n}$ is $a_{n}=2n+1$ without mathematical induction? 
Let  sequence  $a_{1}=3$ and $$a_{n+1}=a^2_{n}-2na_{n}+2,n\ge 1$$
Find the $a_{n}$

My idea: 
since
$$a_{2}=9-6+2=5,a_{3}=a^2_{2}-4a_{2}+2=25-20+2=7,a_{4}=a^2_{3}-6a_{3}+2=49-42+2=9$$
so I inductive
$$a_{n}=2n+1$$
and it is easy use Mathematical induction to prove it
since
$$a_{n+1}=(2n+1)^2-2n\cdot (2n+1)+2=2(n+1)+1$$
My Question,This sequence have without mathematical induction methods?
such as this two  post. At first I thought to can't find other ways, but finally surprised someone gives three different methods:
How find this $n(n+1)a_{n+1}=n(n-1)a_{n}-(n-2)a_{n-1}$
,How find $a_{n}$ if the sequence $a_{n}=2a_{n-1}+(2n-1)^2a_{n-2},n\ge 1$

What you want to prove is: 
$\forall n \in\mathbb{N}: a_n = 2n+1$.
Of course, one can convince oneself that this equation holds by several ways, e.g. by heuristics, by plugging in $2n+1$ into the recursive description or by arguing that some sequence is constant. But in every case one states that some property has to hold for every natural number (and this is obvious since the very claim you want to prove is a claim "for every natural number...").
For me, the most "induction hiding" version is "plugging in $2n+1$ into the recursive description". So I'll take this as an example where induction is hidden:
A proof could look like this:
An example
We have $a_{n+1} = a_n^2 - 2na_n +2$ for $n\geq 1$. Then $a_n = 2n+1$ since if we substitute:
$$\begin{aligned}
a_{n+1} &= (2n+1)^2 -2n(2n+1)+2\\&=4n^2+4n+1+2-4n^2-2n\\&=2n+3\\&=2(n+1)+1
\end{aligned}$$
it works.
Now, in fact, what have we proven? We have proven that:
$$\text{If $a_n = 2n+1$ (for some $n\in\mathbb{N}$) then $a_{n+1} = 2(n+1)+1$ (for this very $n$).}$$
Aha! This is the induction step. But still not the claim in question -- although almost.
