Solving the recurrence with $a_0 = 0$ and $a_n = \left(\frac{n-2}{n}\right)a_{n-1} + 2$

Question

Can anyone here help me to solve this recurrence relation?

$$a_0 = 0,\quad a_n = \left(\frac{n-2}{n}\right)a_{n-1} + 2, \quad n > 0$$

I've tried lots of ways to solve it, unsuccessfully.

Thanks in advance.

Answer 1

Hint: For $n>1$, we have $$a_n=\frac{(n-1)(n-2)}{n(n-1)}a_{n-1}+2,$$ $$n(n-1)a_n=(n-1)(n-2)a_{n-1}+2n(n-1),$$ let $b_n=n(n-1)a_n$ and it is not hard to solve $b_n$ then get $a_n$.