How can you solve this recurrence? How can I solve this recurrence?

$$B_{k}=1+\frac{n-k-1}{n} B_{k+1} + \frac{kx}{n},\qquad x>0$$

This is defined for $1 \leq k \leq n-1$ and $n \geq 2$.  When $k=n-1$ then we can see that $B_{n-1} = 1+ \dfrac{(n-1)x}{n}$ so this is effectively the base case.
 A: First I would rewrite the recursion using 
$C_k = B_{n-1-k}$
s.t. $C_0$ is the base case and we get the recursion
$C_k = 1 + \frac{k}{n}C_{k-1} + \frac{n-k-1}{n}x, \qquad k>0.$
Next one can split $C_k = \frac{D_k}{n^k} + x\frac{E_k}{n^{k+1}}$ into a term with $x$ and a term without $x$, where the denominators $n^k, n^{k+1}$ are for convenience. This gives the recursions
$D_k = kD_{k-1} + n^k, \quad k>0, \quad D_0 = 1$
and
$E_k = k E_{k-1} + (n-k-1)n^k, \quad k>0, \quad E_0=n-1.$
Both recursions can be expanded into simple sums. Thus we get
$$C_k = \frac1{n^k}\sum_{i=0}^k (k)_i n^{k-i} + \frac{x}{n^{k+1}}\sum_0^k \mbox{something similar}$$
where $(k)_i = k(k-1)\ldots (k-i+1)$.
(I hope I got all the indices and exponents right in my quick writing; but the principle should definitely work.)
A: Here is a solution computed by Maple
$$ {\frac { \left( -1 \right) ^{k}{n}^{k} \left( B \left( 0 \right) 
\Gamma  \left( -n+1 \right) -\sum _{{\it m}=0}^{k-1} \left( n+{\it m
}\,x \right) {n}^{-{\it m}-1}\Gamma  \left( -n+{\it m}+1 \right) 
 \left( -1 \right) ^{{\it m}} \right) }{\Gamma  \left(-n+k+1 \right) }},$$
where $ \Gamma(s) $ is the Gamma function.
