If t1 = 6, t2 = 4, t3 = 2, and tn = (tn-1 + tn-2) x tn-1, find t7.

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- Jan 4th 2009, 10:58 AMcasey_kSequences
If t1 = 6, t2 = 4, t3 = 2, and tn = (tn-1 + tn-2) x tn-1, find t7.

- Jan 4th 2009, 11:26 AMearboth
- Jan 4th 2009, 11:36 AMcasey_kRe:
the sequence is 6, 4, 2

the recursive formula is tn = (tn-3 + tn-2) x tn-1

so I did it = (2+4) x 6 = 36

and i got 2,068, 416 for t7

but it's not right and now i'm not sure how to solve it. (Doh) - Jan 4th 2009, 12:40 PMskeeter
- Jan 4th 2009, 12:42 PMcasey_kFormula
$\displaystyle

t_n = (t_{n-3} + t_{n-2}) \cdot t_{n-1}

$

Sorry I didn't post my question clearly, still getting the hang of codes. - Jan 5th 2009, 02:06 AMearboth
Using this equation I've got:

$\displaystyle n = 4:\left\{\begin{array}{lcr}t_4&=&(t_1+t_2)\cdot t_3 \\ &=&(6+4) \cdot 2 = \boxed{20}\end{array}\right.$

$\displaystyle n = 5:\left\{\begin{array}{lcr}t_5&=&(t_2+t_3)\cdot t_4 \\ &=&(4+2) \cdot 20 = \boxed{120}\end{array}\right.$

$\displaystyle n = 6:\left\{\begin{array}{lcr}t_6&=&(t_3+t_4)\cdot t_5 \\ &=&(2+20) \cdot 120 = \boxed{2640}\end{array}\right.$

$\displaystyle n = 7:\left\{\begin{array}{lcr}t_7&=&(t_4+t_5)\cdot t_6 \\ &=&(20+120) \cdot 2640 = \boxed{369600}\end{array}\right.$