Answer
$\displaystyle a_1=1$
$\displaystyle a_2=3$
$\displaystyle a_3=15$
$\displaystyle a_4=105$
$\displaystyle a_{10}=654729075$
Work Step by Step
We are given:
$a_{n}= \frac{(2n)!}{2^{n}n!}$
We evaluate:
$\displaystyle a_1 = \frac{(2*1)!}{2^{1}1!}=\frac{2}{2}=1$
$\displaystyle a_2= \frac{(2*2)!}{2^{2}2!}=\frac{4*3*2!}{4*2!}=3$
$\displaystyle a_3= \frac{(2*3)!}{2^{3}3!}=\frac{6*5*4*3!}{8*3!}=15$
$\displaystyle a_4= \frac{(2*4)!}{2^{4}4!}=\frac{8*7*6*5*4!}{16*4!}=105$
$\displaystyle a_{10}= \frac{(2*10)!}{2^{10}*10!}=\frac{20!}{2^{10}*10!}=654729075$
