# $n^{2}T(n)=(n-1)(n-2)T(n-1)+3$ with initial condition $T(1) = 1$

I find that the above nonlinear recurrence could be written as the following $$n^{2}T(n)=(n-1)^{2}T(n-1)-(n-1)T(n-1)+3$$. Then, I let $$\alpha(n) = nT(n-1)$$. However, I am stuck here. I don't know how to solve it. Hope that someone could give me some hint about it.

I think you made a type, however defining $$\alpha(n) := n T(n)$$ leads to the following expression: $$n \alpha(n) = (n-2)\alpha(n-1) + 3.$$ We know that $$\alpha(1) = 1$$, then from this we get $$\alpha(2) = 3/2$$ and we notice that it is true for all $$n \ge 2$$. To prove this we use induction.
Assume it is true for $$n - 1$$, i.e. $$\alpha(n - 1) = 3/2$$ then $$n \alpha(n) = (n - 2) \cdot 3/2 + 3 = n \cdot 3/2 \implies \alpha(n) = 3/2,$$ concluding the proof by induction. Hence, $$\alpha(n) = 3/2$$ for all $$n \ge 2$$.
Finally, $$T(n) = \frac{3}{2n} ~ \forall n \ge 2.$$