# Sequences

• Jan 4th 2009, 11:58 AM
casey_k
Sequences
If t1 = 6, t2 = 4, t3 = 2, and tn = (tn-1 + tn-2) x tn-1, find t7.
• Jan 4th 2009, 12:26 PM
earboth
Quote:

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

Are you sure you typed the question correctly?

According to your recursive equation you get:

$t_3 = (t_2+t_1)\cdot t_2 = (4+6)\cdot 4 = \boxed{40 \neq 2}$
• Jan 4th 2009, 12:36 PM
casey_k
Re:
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, 01:40 PM
skeeter
Quote:

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

I'm guessing here, but is this your equation?

$t_n = (t_{n-1} + t_{n-2}) \cdot t_{n-1}$
• Jan 4th 2009, 01:42 PM
casey_k
Formula
$

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, 03:06 AM
earboth
Quote:

Originally Posted by casey_k
$

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.

Using this equation I've got:

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

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

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

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