Hi,(adsbygoogle = window.adsbygoogle || []).push({});

I'm a bit confused when it comes to the simplification here. I must've had a brain fart or something when we were going over this, but if I had the following series:

[tex]\sum _{n=1} ^{\infty} 2^{2n}3^{1-n}[/tex]

I know enough to go this far:

[tex]\sum _{n=1} ^{\infty} \frac{2^{2n}}{3^{n-1}}[/tex]

But my text book goes on to make the above into this:

[tex]\sum _{n=1} ^{\infty} 4(\frac{4}{3})^{n-1}[/tex]

I know that this is just a simplification, but how do they get from my second step to the above step? I thought I knew how, but this proved wrong on a couple of different problems I was working on.

Also, for this series:

[tex]\sum _{n=1} ^{\infty} \frac{3^n+2^n}{6^n}[/tex]

I've done this:

[tex]\sum _{n=1} ^{\infty} \frac{3^n+2^n}{6^n} = \sum _{n=1} ^{\infty} \frac{3^n}{6^n}+\sum _{n=1} ^{\infty} \frac{2^n}{6^n} = \sum _{n=1} ^{\infty} (\frac{1}{2})^n + \sum _{n=1} ^{\infty} (\frac{1}{3})^n[/tex]

So for the first sum, a=1 and r=1/2, and for the second sum a=1 and r=1/3.

But this is apparently wrong. From the solution manual, for the first one, a=1/2 and r=1/2, for the second one, a=1/3 and r=1/3.

Can someone explain this?

Thanks!

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Series Problem and Simplification

**Physics Forums | Science Articles, Homework Help, Discussion**