Answer
$\left\{\begin{array}{l}
a_{1}=1,\ a_{2}=2,\\
a_{n}=a_{n-1}\cdot a_{n-2}
\end{array}\right.$
Work Step by Step
Note that, starting from the third term (n=3)
the current (nth) term equals the product of the preceding two.
$ a_1=1,\ \ a_2=2,$
$a_{3}\quad=2=2\times 1\quad=a_{2}\cdot a_{1}=a_{3-1}\cdot a_{3-2}$
$a_{4}\quad=4=2\times 2\quad=a_{3}\cdot a_{2}=a_{4-1}\cdot a_{4-2}$
$a_{5}\quad=8=4\times 2\quad=a_{4}\cdot a_{3}=a_{5-1}\cdot a_{5-2}$
$a_{6}\quad=32=8\times 4\quad=a_{5}\cdot a_{4}=a_{6-1}\cdot a_{6-2}$
From this pattern, $a_{n}=a_{n-1}\cdot a_{n-2}$, for $n\geq 3$.
For a recursive rule, we give the information about the first terms
and a rule on how to obtain the next term from the preceding ones.
Here,
$\left\{\begin{array}{l}
a_{1}=1,\ a_{2}=2,\\
a_{n}=a_{n-1}\cdot a_{n-2}
\end{array}\right.$
