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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!

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# Series Problem and Simplification

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