I'm trying to express the following recursive relationship as a summation \begin{align} a_2 &= \frac{-\alpha(\alpha+1)}{2!}a_0 \\ a_4 &= \frac{\alpha(\alpha+1)(\alpha-2)(\alpha+3)}{4!}a_0 \\ a_6 &= \frac{-\alpha(\alpha+1)(\alpha-2)(\alpha+3)(\alpha-4)(\alpha+5)}{6!}a_0\\ & \ \ \vdots \end{align}
Attempt
Genouinely confused , I suppose it should look something along those lines $$ a_{2n} = \sum_{n\ge0}\frac{(-1)^n (\alpha+1)\dots(\alpha +(2n-1))}{(2n)!}a_0,$$ but I'm not sure if something should go in between the two factors in the numerator.