Task. Let $A(t)$ and $B(t)$ be generating functions of sequences ${a_n} = (-1)^n C_{2n}^n 4^{-n}$ and ${b_n}$. Let $A(t)B(t) = 1$. Find $b_n$ and $B(t)$.
I try do next: $A(t)B(t) = 1 =>$ $$ \begin{cases} a_0b_0 = 1 \\ a_0b_1 + a_1b_0 = 0 \\ a_0 b_2 + a_1b_1 + a_2 b_0 = 0 \\ ... \\ \sum_{k=0}^n a_k b_{n - k} = 0 \end{cases} $$
\begin{cases} b_0 = 1/ a_0 = 1 \\ b_1 = -a_1\\ b_2 = - a_1b_1 - a_2 b_0 \\ ... \\ b_n = -\sum_{k=1}^n a_k b_{n - k} \end{cases}
I try to find explicit formula for $b_n$, but I have no idea how to do it. Could u help me, please?
Writing $\binom{2n}{n}$ for $C_{2n}^n$, we have $$A(t) = \sum_{n \ge 0} a_n t^n = \sum_{n \ge 0} \binom{2n}{n} (-t/4)^n = \frac{1}{\sqrt{1+t}},$$ so $$B(t) = \frac{1}{A(t)} = \sqrt{1+t} = \sum_{n \ge 0} \binom{1/2}{n} t^n,$$ yielding $b_n = \binom{1/2}{n}$.