Answer
\begin{array}{lclcl}
b_1=1&\\
b_2=3&\\
b_3=20&\\
b_4=210\\
\end{array}
Work Step by Step
Given $$ b_{n}=\frac{(2 n-1) !}{n !}, \ \ \ \ \ \ n= 1,2,3, \dots
$$
So, we get
\begin{array}{|l|l|}\hline
& n& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ b_n \\ \\ \hline
\text{First Term} & 1 & b_1= \frac{(2 \times 1-1) !}{1 !}=\frac{1 !}{1}=1 \\ \\ \hline
\text{Second Term} & 2 & b_2=\frac{(2 \times 2-1) !}{2 !}=\frac{3 !}{2 !}=\frac{3 \times2 !}{2 !}=3 \\ \\ \hline
\text{Third Term} & 3 & b_3= \frac{(2 \times 3-1) !}{3 !}=\frac{5 !}{3 !}=\frac{5 \times 4 \times 3 !}{3 !}=20 \\ \\ \hline
\text{Fourth Term} & 4 & b_4= \frac{(2 \times 4-1) !}{4 !}=\frac{7 !}{4 !}=\frac{7 \times 6 \times \ 5 \times 4 !}{4 !}=210\\ \\ \hline
\end{array}
