what is..... the value?

**bobbluecow** · October 20th 2006, 10:59 AM

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!

**Jameson** · October 20th 2006, 11:07 AM

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?

**topsquark** · October 20th 2006, 11:08 AM

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

**Jameson** · October 20th 2006, 11:11 AM

Sigh. My help again is useless.

**topsquark** · October 20th 2006, 11:12 AM

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

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