1. ## 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.

2. ## Re: Sequences and Series

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 $\displaystyle a^{b+c} = a^ba^c$ to express $\displaystyle 3^{n+1}$ in terms of $\displaystyle 3^n$

Spoiler:
$\displaystyle \dfrac{3^{n+1} - 3^n}{2} = \dfrac{3(3^n) - 3^n}{2} = \dfrac{3^n(3-1)}{2} = 3^n$

3. ## Re: Sequences and Series

$\displaystyle \frac{3^{n+1}}{2} - \frac{3^n}{2} = \frac{3 \cdot 3^n}{2} - \frac{1 \cdot 3^n}{2} = \frac{(3 - 1) \cdot 3^n}{2} = ...$

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

4. ## Re: Sequences and Series

Thanks for the help guys. I really appreciate it.