anonymous
  • anonymous
By looking at 0.36 (repeating decimal) like an infinite geometric sequence, convert it into a fraction.
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
you can do this a couple ways
anonymous
  • anonymous
is it \(\overline{.36}\) ?
anonymous
  • anonymous
Yes.

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anonymous
  • anonymous
I have the formula \[S _{n}=\frac{ a _{1}-a_{1}r ^{n} }{ 1-r }\] If that helps.
anonymous
  • anonymous
ok one simple method is to call \(x=\overline{.36}\) so \(100x=36\overline{.36}\) subtract and get \(100x-x=36\) so \[99x=36\] and therefore \[x=\frac{36}{99}=\frac{4}{11}\] but that might not be what you want
anonymous
  • anonymous
@satellite73 hey, when you get a chance can you finish helping me with that problem we started? thanks :)
anonymous
  • anonymous
if you want to use the formula you wrote above, then since you are summing an infinite geometric series omit the \(-a_1r^n\) part and go with \[S=\frac{a_1}{1-r}\]
anonymous
  • anonymous
Oh. I think I copied the formula down incorrectly...
anonymous
  • anonymous
Cuz I was looking through my notes and wondering how I got the answers with the -a1r^n part
anonymous
  • anonymous
yeah since \(r<1\) you have \(\lim_{n\to \infty}r^n=0\)
anonymous
  • anonymous
Thanks.
anonymous
  • anonymous
in this case you can use \[a=\frac{36}{100}\] and \[r=\frac{1}{100}\]
anonymous
  • anonymous
For reference, though, does that formula equal\[S_{n}= \frac{ a_{1}(1-r ^{n)} }{ 1-r }\]
anonymous
  • anonymous
you get \[S=\overline{.36}=\frac{\frac{36}{100}}{1-\frac{1}{100}}\] \[=\frac{\frac{36}{100}}{\frac{99}{100}}\] \[=\frac{36}{99}\] etc
anonymous
  • anonymous
no that is a formula for a finite series infinite series is the one i wrote above
anonymous
  • anonymous
Ah. Well, thanks!
anonymous
  • anonymous
@erin512 repost i cannot find it

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