How to solve this recursion relation? Suppose:
$$2k(n-k)a_k=n(n-1)+(n-k)(k+1)a_{k+1}+(n-k)(k-1)a_{k-1}$$
where $k=1, 2, ..., n-1$ and $a_n=0$,
how to derive $a_k$?
I tried to find pattern $a_1-a_2=n/2$; $a_2-a3=n(2n-1)/6(n-2)$, it become more and more complicated and I can't find the rule.

Hint.
Making $b_k = k a_k$ we have
$$
-b_{k-1}+2 b_k - b_{k+1} = \frac{n(n+1)}{n-k}
$$
This can be simplified by making $c_k = b_k-b_{k-1}$ giving
$$
c_k-c_{k+1} = \frac{n(n+1)}{n-k}
$$
After solving for $c_k$ can be then solved for $b_k$
