I am trying to find the taylor series expansion about $0$ (maclaurin series) of $$x \rightarrow \frac{1}{\sqrt{1-\beta x(x+1)}} \text{ with } \beta \in \mathbb{R}^{+*}$$
I've tried using the taylor series expansion of $$\frac{1}{\sqrt{1-X}} = \sum_{n=0}^{\infty}4^{-n}{2n \choose n}X^n \text{ }\text{ }\text{ }\text{ with } \text{ } X=\beta x(x+1)$$
But I can't turn it into a power series because of the $(x+1)^n$...
I've also tried to derive $$\frac{1}{n!}\cdot\frac{\text{d}^n}{\text{d}x^n}\left(\frac{1}{\sqrt{1-\beta x(x+1)}}\right)_{x=0}$$ But no results so far...
Edit : with a more powerful method, I found that, if we call $\left(a_n\right)_{n\in\mathbb{N}}$ the coefficients of the taylor series expansion $\left(\frac{1}{\sqrt{1-\beta x(x+1)}}=\sum_{n=0}^{\infty}a_nx^n\right)$, the sequence $\left(a_n\right)_{n\in\mathbb{N}}$ is then defined by : $$\forall n\geq3, \text{} na_n=\beta\left(n-\frac{1}{2}\right)a_{n-1}+\beta\left(n-1\right)a_{n-2}$$ $$\text{with }\text{ }a_1 = \frac{\beta}{2} \text{ , } a_2 = \frac{3}{8}\beta^2+\frac{1}{2}\beta$$
It's definitely a step forward, but I don't know how to proceed from there. Is there a way to handle sequences that are defined by such a way ?