Answer
$$c_1=3, \quad c_2= \frac{9}{2}, \quad c_3= \frac{9}{2}, \quad c_4= \frac{27}{8} .$$
Work Step by Step
Since $c_n=\frac{3^n}{ n!}$, then we have
$$c_1=\frac{3}{ 1!}=3, \quad c_2=\frac{3^2}{ 2!}=\frac{9}{2}, \quad c_3=\frac{3^3}{ 3!}=\frac{27}{6}=\frac{9}{2}, \quad c_4=\frac{3^4}{ 4!}=\frac{81}{24}=\frac{27}{8} .$$
