1. 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!

2. 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?

3. 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

4. Sigh. My help again is useless.

5. Originally Posted by Jameson
Sigh. My help again is useless.
Not useless. In fact it was MY help that was useless since you posted first.

-Dan