How show that $a_{n}=n$ if $ a_{n+1}+a_{n-1}=\frac{2n}{a_{n}-a_{n-1}}$ define sequence $\{a_{n}\}$ such $a_{1}=1,a_{2}=2$, and such
$$ a_{n+1}+a_{n-1}=\dfrac{2n}{a_{n}-a_{n-1}},n\ge 2$$
show that:$$a_{n}=n$$
I want use  without induction solve this sequence?
 A: Yes. Use induction. Suppose that $a_{k}=k$ for $k=1,...,n$. Then $$a_{n+1}=\frac{2n}{a_{n}-a_{n-1}}-a_{n-1}=\frac{2n}{n-(n-1)}-(n-1)=2n-n+1=n+1.$$
A: There might be no way to solve this without induction. I will prove this at least for attempts making use of any kind of algebraic or analytic manipulation. For attempts making use of construction (or something else), this may or may not apply.
Proof
The recurrence relation at hand requires that to evaluate any general $a_{n}, n \gt 2$, we know the difference $a_{n - 1} - a_{n - 2}$, beforehand. For any such general $n$, this difference might be $0$. To attempt to conduct any kind of analysis on $a_n$ must then assume that this difference can never be $0$. However, without having conducted any kind of analysis on $a_n$, this assumption is not justified. This, in fact, is exactly what the inductive step accomplishes. Therefore, no algebraic or analytic proof not making use of induction exists for this recurrence relation's closed form.
Now, I suspect induction might somehow show up for other approaches to this problem as well, perhaps not directly related to $a_n$. However, I cannot say anything about these cases with certainty.
