# Sequences and Series

• October 14th 2011, 10:05 AM
GrigOrig99
Sequences and Series

In this case, the text book has included the answer. I have 3^n/2, they have just 3^n. Can someone help me figure out how to factor out the 2? Thank you.
• October 14th 2011, 10:11 AM
e^(i*pi)
Re: Sequences and Series
Quote:

Originally Posted by GrigOrig99

In this case, the text book has included the answer. I have 3^n/2, they have just 3^n. Can someone help me figure out how to factor out the 2? Thank you.

You're very close, it is that last simplification that is wrong - the laws of exponents do not work like that. Use the law that says $a^{b+c} = a^ba^c$ to express $3^{n+1}$ in terms of $3^n$

Spoiler:
$\dfrac{3^{n+1} - 3^n}{2} = \dfrac{3(3^n) - 3^n}{2} = \dfrac{3^n(3-1)}{2} = 3^n$
• October 14th 2011, 10:12 AM
TheChaz
Re: Sequences and Series
$\frac{3^{n+1}}{2} - \frac{3^n}{2} = \frac{3 \cdot 3^n}{2} - \frac{1 \cdot 3^n}{2} = \frac{(3 - 1) \cdot 3^n}{2} = ...$

edit: the biggest spoiler was that someone posted before me!

(Punch)
• October 14th 2011, 11:03 AM
GrigOrig99
Re: Sequences and Series
Thanks for the help guys. I really appreciate it.