How to solve this recursion relation?

Question

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.

Answer 1

We can finish Cesareo’s solution as follows. For each $1\le k\le n$ we have $c_k=c_{n}+n(n+1)\sum_{i=1}^{n-k}\frac 1i$. Then for each $1\le l\le n$,

$$b_{l-1}=-\sum_{k=l}^n c_k=-\sum_{k=l}^n c_{n}+n(n+1)\sum_{i=1}^{n-k}\frac 1i=$$ $$- (n-l+1)c_n- n(n+1)\sum_{k=l}^n\sum_{i=1}^{n-k}\frac 1i=$$ $$- (n-l+1)c_n- n(n+1)\sum_{i=1}^{n-l}\frac{n-i-l+1}i=$$ $$- (n-l+1)c_n- n(n+1)\left(l-n+(n-l+1)\sum_{i=1}^{n-l}\frac{1}i\right).$$

Remark that

$$0=b_0=- nc_n- n(n+1)\left(1-n+n\sum_{i=1}^{n-1}\frac{1}i\right).$$

We assume that $n\ge 1$. Then

$$c_n=-(n+1)\left(1-n+n\sum_{i=1}^{n-1}\frac{1}i\right),$$

so for $2\le l\le n,$

$$a_{l-1}=\frac {b_{l-1}}{l-1}=\frac{n+1}{l-1}\left((n-l+1)\left(1-n+n\sum_{i=1}^{n-1}\frac{1}i\right)-n\left(l-n+(n-l+1)\sum_{i=1}^{n-l}\frac{1}i\right)\right)=$$ $$\frac{n+1}{l-1}\left(1-l+n(n-l+1) \sum_{i=n-l+1}^{n-1}\frac{1}i \right).$$

Finally, for $1\le k\le n-1,$

$$a_{k}=\frac{n+1}{k}\left(-k+n(n-k) \sum_{i=n-k}^{n-1}\frac{1}i \right).$$

Answer 2

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 $b_k$