# what is..... the value?

• October 20th 2006, 11:59 AM
bobbluecow
what is..... the value?
okay the question is as follows:
write the first three terms of tn= 5(2)^ n-1. what is the value of
tn/tn-3?

okay so the first three terms are as follows:
t1= 5(2)^1-1=5X1=5
t2= 5(2)^2-1=5X 2=10
t3= 5(2)^3-1=5X4= 20

okay I am a little bit confused about the next part: what is the value of
tn/tn-3? how would I do this? thanks for your help!
• October 20th 2006, 12:07 PM
Jameson
Ok, from your information $t_n=5 \times 2^{n-1}$. To get $t_{n-3}$ just substitute $n-3$ for $n$. Thus $t_{n-3}=5 \times 2^{(n-3)-1}=5 \times 2^{n-4}$.

Now divide the two. Can you see the simplification?
• October 20th 2006, 12:08 PM
topsquark
Quote:

Originally Posted by bobbluecow
okay the question is as follows:
write the first three terms of tn= 5(2)^ n-1. what is the value of
tn/tn-3?

okay so the first three terms are as follows:
t1= 5(2)^1-1=5X1=5
t2= 5(2)^2-1=5X 2=10
t3= 5(2)^3-1=5X4= 20

okay I am a little bit confused about the next part: what is the value of
tn/tn-3? how would I do this? thanks for your help!

$t_n = 5 \left ( 2^{n-1} \right )$

So
$t_{n-3} = 5 \left ( 2^{n-4} \right )$

Thus
$\frac{t_n}{t_{n-3}} = \frac{5 \left ( 2^{n-1} \right )}{5 \left ( 2^{n-4} \right )}$ = $\frac{2^{n-1}}{2^{n-4}} = 2^{(n-1)-(n-4)} = 2^{-1+4} = 2^3 = 8$

-Dan
• October 20th 2006, 12:11 PM
Jameson
Sigh. My help again is useless. :rolleyes:
• October 20th 2006, 12:12 PM
topsquark
Quote:

Originally Posted by Jameson
Sigh. My help again is useless. :rolleyes:

Not useless. In fact it was MY help that was useless since you posted first. :)

-Dan