anonymous
  • anonymous
If this calculation continued forever, what would you expect the answer to be: 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + ...
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...\] so you know how to add this up ?
anonymous
  • anonymous
no i dont
anonymous
  • anonymous
\[a+ar+ar^2+ar^3+ar^4+...=\frac{a}{1-r}\] use \[a=r=\frac{1}{3}\]

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More answers

anonymous
  • anonymous
you can almost do it in your head almost what is one minus one third?
anonymous
  • anonymous
2/3
anonymous
  • anonymous
ok good what is the reciprocal of \(\frac{2}{3}\) ?
anonymous
  • anonymous
3/2
anonymous
  • anonymous
and finally, what is \(\frac{3}{2}\times \frac{1}{3}\) ?
anonymous
  • anonymous
1/2
anonymous
  • anonymous
you win
anonymous
  • anonymous
im trying to understand the formula how it works
anonymous
  • anonymous
the formula i wrote? \[\frac{a}{1-r}\]?
anonymous
  • anonymous
yes
anonymous
  • anonymous
you mean what to plug in for \(a\) and \(r\)? or do you mean why the formula works? or do you mean how you compute using the formula?
anonymous
  • anonymous
what is a and what is r?
anonymous
  • anonymous
\(a\) is the first term, and \(r\) is what you multiply the first term by to get the second term what you multiply the second term by to get the third term, and so on
anonymous
  • anonymous
this formula only works if you have a sum that looks like \[a+ar+ar^2+ar^3+ar^4+...\]
anonymous
  • anonymous
in your example you do you have \[\frac{1}{3}+\frac{1}{3}\times \frac{1}{3}+\frac{1}{3}\times \left(\frac{1}{3}\right)^2+...\]
anonymous
  • anonymous
in other words, you start with \(\frac{1}{3}\) as the first term and keep multiplying each successive term by \(\frac{1}{3}\) it is not always the case that \(a=r\) but it is in your example
anonymous
  • anonymous
perfectly explained thanks alot @satellite73
anonymous
  • anonymous
yw
anonymous
  • anonymous
on e last question: how did you deduct the formula?
anonymous
  • anonymous
what if it is 1/3 + 1/9 - 1/27 + 1/81 - 1/243 + ...
anonymous
  • anonymous
@satellite73
anonymous
  • anonymous
hmm
anonymous
  • anonymous
drop off the first \(\frac{1}{3}\) and it looks like \(a=\frac{1}{9}\) and \(r=-\frac{1}{3}\)
anonymous
  • anonymous
compute \[\frac{\frac{1}{9}}{1+\frac{1}{3}}\] and then add \(\frac{1}{3}\) to the result
anonymous
  • anonymous
could you please explain the formula?
anonymous
  • anonymous
why is a =1/9
anonymous
  • anonymous
1 - 1/2 + 1/4 - 1/8 + 1/16 - 1/32 + 1/64 - ...
anonymous
  • anonymous
this is the problem i cant get
anonymous
  • anonymous
you still there? i was off on a tangent
anonymous
  • anonymous
\[1 - 1/2 + 1/4 - 1/8 + 1/16 - 1/32 + 1/64... \] it alternates so \(r\) is negative
anonymous
  • anonymous
\(a=1,r=-\frac{1}{2}\)
anonymous
  • anonymous
compute as before \[\frac{a}{1-r}\] with \(a=1,r=-\frac{1}{2}\) to get \(\frac{1}{1+\frac{1}{2}}\)
anonymous
  • anonymous
why is a=1 and r=-1/2?
anonymous
  • anonymous
but does the same logic apply for the first question @satellite73
anonymous
  • anonymous
1/3 + 1/9 + 1/27 + 1/81 + 1/243 + ...
anonymous
  • anonymous
will 1/3 be the first term (a=1/3)???
anonymous
  • anonymous
what will r be? 1st term multiply 2nd term (1/3 x 1/9)
anonymous
  • anonymous
@satellite73

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