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you can do this a couple ways

is it \(\overline{.36}\) ?

Yes.

I have the formula \[S _{n}=\frac{ a _{1}-a_{1}r ^{n} }{ 1-r }\]
If that helps.

Oh. I think I copied the formula down incorrectly...

Cuz I was looking through my notes and wondering how I got the answers with the -a1r^n part

yeah since \(r<1\) you have \(\lim_{n\to \infty}r^n=0\)

Thanks.

in this case you can use
\[a=\frac{36}{100}\] and \[r=\frac{1}{100}\]

For reference, though, does that formula equal\[S_{n}= \frac{ a_{1}(1-r ^{n)} }{ 1-r }\]

no that is a formula for a finite series
infinite series is the one i wrote above

Ah. Well, thanks!