Finding an Explicit Formula for $a_n$ from a Recurrence Relation I have the following recurrence relation
$$\beta ^n n(n+2)a_n = \sum_{k=0}^{n-1}  \beta^k (\alpha+k+1) a_k  , \quad  a_0=1, \quad n \ge 1 \tag{1}$$
where $\beta$ and $\alpha$ are positive real numbers.
I know that one can easily find $a_n$ by back substitution. Here is my question

Is it possible to find an explicit formula for $a_n$?


My Work
I just computed the first three terms to see what I can guess. But I could not come up with anything!
$$\begin{align}
n&=1, \quad a_1=\beta^{-1} \left[ \frac{\alpha + 1}{1 \times 3} \right] \\
n&=2, \quad a_2=\beta^{-2}\left[ \frac{\alpha+1}{2 \times 4} + \frac{(\alpha+1)(\alpha+2)}{(1 \times 3)(2 \times 4)} \right] \\
n&=3, \quad a_3=\beta^{-3}\left[ \frac{\alpha+1}{3 \times 5}+\frac{(\alpha+1)(\alpha+2)}{(1 \times 3)(3 \times 5)}+\frac{(\alpha+1)(\alpha+3)}{(2 \times 4)(3 \times 5)}+\frac{(\alpha+1)(\alpha+2)(\alpha+3)}{(1 \times 3)(2 \times 4)(3 \times 5)} \right]
\end{align}$$

Inspired by the comment of Achille Hui, I noticed that
$$\begin{align}
\beta ^n n(n+2)a_n &= \sum_{k=0}^{n-1}  \beta^k (\alpha+k+1) a_k \\
\beta ^{n-1} (n-1)(n+1)a_{n-1} &= \sum_{k=0}^{n-2}  \beta^k (\alpha+k+1) a_k 
\end{align}$$
and subtracting the above equations will lead to
$$\begin{align}
\beta ^n n(n+2)a_n - \beta ^{n-1} (n-1)(n+1)a_{n-1} &= \beta^{n-1}(\alpha+n)a_{n-1} \\
\beta  n(n+2)a_n &= [n^2+n+(\alpha-1)] a_{n-1}
\end{align}$$
and finally
$$a_n=\frac{[n-(\frac{-1+\sqrt{5-4\alpha}}{2})][n-(\frac{-1-\sqrt{5-4\alpha}}{2})]}{\beta  n(n+2)}a_{n-1}$$
one can easily solve the above recurrence relation to get
$$a_n = \frac{\left(\frac{3+\sqrt{5-4\alpha}}{2}\right)_n \left( \frac{3-\sqrt{5-4\alpha}}{2}\right)_n}{\beta^n n! (3)_n}$$
where $(x)_n$ is the Pochhammer symbol.
