If t1 = 6, t2 = 4, t3 = 2, and tn = (tn-1 + tn-2) x tn-1, find t7.
Using this equation I've got:
$\displaystyle n = 4:\left\{\begin{array}{lcr}t_4&=&(t_1+t_2)\cdot t_3 \\ &=&(6+4) \cdot 2 = \boxed{20}\end{array}\right.$
$\displaystyle n = 5:\left\{\begin{array}{lcr}t_5&=&(t_2+t_3)\cdot t_4 \\ &=&(4+2) \cdot 20 = \boxed{120}\end{array}\right.$
$\displaystyle n = 6:\left\{\begin{array}{lcr}t_6&=&(t_3+t_4)\cdot t_5 \\ &=&(2+20) \cdot 120 = \boxed{2640}\end{array}\right.$
$\displaystyle n = 7:\left\{\begin{array}{lcr}t_7&=&(t_4+t_5)\cdot t_6 \\ &=&(20+120) \cdot 2640 = \boxed{369600}\end{array}\right.$