[tex]\text{The explicit rule of geometric sequence}\\\\a_n=a_1 r^{n-1}\\------------------------------\\\text{We have the recursive form}\ a_1=6,\ a_n=2\cdot a_{n-1}.\\\\\text{Therefore}\ r=2.\ \text{Substitute:}\\\\a_n=6\cdot2^{n-1}=6\cdot2^n\cdot2^{-1}=6\cdot2^n\cdot\dfrac{1}{2}=\boxed{3\cdot2^n}[/tex]
