Algebra 2 (1st Edition)

$\left\{\begin{array}{l} a_{1}=1,\ a_{2}=2,\\ a_{n}=a_{n-1}\cdot a_{n-2} \end{array}\right.$
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, $\left\{\begin{array}{l} a_{1}=1,\ a_{2}=2,\\ a_{n}=a_{n-1}\cdot a_{n-2} \end{array}\right.$