I'm trying to come up with a generating function for $\{(n+2)C_{n+1}\}^\infty_{n=0}$ where $C_n$ is the $n$th Catalan number.
I know we can write $(n+2)C_{n+1} = 2(2n+1)C_n$. I also tried to follow this post: Simplifying Catalan number recurrence relation
However I am getting somewhat stuck because of the added $(n+2)$ term.
Note that $$(n+1)C_n=\binom{2n}{n}\qquad\qquad n\geq 0$$ are the Central Binomial Coefficients with the generating series representation \begin{align*} \sum_{n=0}^\infty\binom{2n}{n}x^n=\frac{1}{\sqrt{1-4x}}\qquad\qquad |4x|<1\tag{1} \end{align*}
A generating function for $\{(n+2)C_{n+1}\}^\infty_{n=0}$ is therefore
\begin{align*} \sum_{n= 0}^\infty& (n+2)C_{n+1}x^n\\ &=\sum_{n=1}^\infty (n+1)C_{n}x^{n-1}\\ &=\frac{1}{x}\sum_{n=1}^\infty \binom{2n}{n}x^{n}\\ &=\frac{1}{x}\left(\frac{1}{\sqrt{1-4x}}-1\right) \end{align*}
Hint: The representation (1) is an application of the binomial series \begin{align*} (1+x)^{\alpha}=\sum_{n=0}^{\infty}\binom{\alpha}{n}x^n\qquad |x|<1, \alpha\in\mathbb{C} \end{align*} and the relation \begin{align*} \binom{-\frac{1}{2}}{n}=\frac{(-1)^n}{4^n}\binom{2n}{n} \end{align*}
