anonymous
  • anonymous
find the sum of the finite geometric series.... 4+ 4/3+ 4/9+ 4/27+...+ 4/6561 i don't know if i'm using the correct formula?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I'm using: sn=\[\frac{ 1-r^n }{ 1-r }\]
RadEn
  • RadEn
then times a1, with a1 is the first term
AkashdeepDeb
  • AkashdeepDeb
@morningperson1991 Is your doubt cleared? :)

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More answers

anonymous
  • anonymous
oops! i forgot to add that to the formula i put up. it still looks like an unbelievably small number. my r=1/3 and my a1=4. i tried to find n by using the formula: a1*\[r^n-1\] but i'm stuck in that step. suggestions?
anonymous
  • anonymous
I guess the trouble is with finding n..my answer is extremely small
AkashdeepDeb
  • AkashdeepDeb
Use ar^n-1 = 4/6561 a = 4 r = 1/3 Find n like that. :)
anonymous
  • anonymous
i believe thats the same formula i've been using @AkashdeepDeb :/ i'll try again now
AkashdeepDeb
  • AkashdeepDeb
Hahah...Yeah :D
anonymous
  • anonymous
since i'm multiplying 4 by the r term. should i divide 4 by each side to get rid of it? my calculator is giving me crazy numbers that don't turn into fractions...
AkashdeepDeb
  • AkashdeepDeb
Hold on let me try it.
anonymous
  • anonymous
@AkashdeepDeb are you getting the (1/3)^n-1 = 1.5241E-4 ? this number is crazy.. i can't turn it into a fraction
AkashdeepDeb
  • AkashdeepDeb
hold on hold on :P
anonymous
  • anonymous
@akashdeepDeb do you know of another formula i can use to find n?
phi
  • phi
what formula are you using to find n ?
anonymous
  • anonymous
I'm using \[a _{1}r^n-1=\frac{ 4 }{ 6561 }\]
phi
  • phi
you mean \[ a _{1}r^{n-1}=\frac{ 4 }{ 6561 } \]
anonymous
  • anonymous
oh yes! i've been using that formula. that was a typo while i input the formula. sorry. I'm just stuck on the n part... i get crazy small numbers that i cant turn into fractions. any suggestions? @phi
phi
  • phi
with a1= 4 you get \[ 4r^{n-1}=\frac{ 4 }{ 6561 } \\ \left(\frac{1}{3}\right)^{n-1}=\frac{ 1 }{ 6561 }\]
phi
  • phi
we can flip both sides, just so it looks nicer \[ 3^{n-1}= 6561\] How would you solve this? I would use logs: \[ (n-1) \log(3) = \log(6561) \\ (n-1) = \frac{\log(6561)}{\log(3)} \\ n = \frac{\log(6561)}{\log(3)}+1\]
anonymous
  • anonymous
@phi that looks very understandable except for the first step.. when you divided both sides by 4. how did it not change the denominator? it only divided by the numerator. can you please explain? sorry, im trying to understand this problem :/ Thank you so much for those steps!
phi
  • phi
are you asking about \[ 4r^{n-1}=\frac{ 4 }{ 6561 } \] ? multiply both sides by 1/4 (or divide both sides by 4) you get \[ r^{n-1}=\frac{ \cancel{4} }{ 6561 }\cdot \frac{1}{\cancel{4}} = \frac{ 1 }{ 6561 } \]
AkashdeepDeb
  • AkashdeepDeb
Sorry I was out.
anonymous
  • anonymous
ohhh! that's embarrassing... @phi thank you so much! Just to verify. can your calculator turn the final sn answer into fractions? because my couldn't. Once i put in n=9 into the geometric summation formula i got 5.99969. I don't know if my teacher will allow me to round that up to 6. but just to make sure, did you get the same answer? @AkashdeepDeb , that's alright :) thank you for trying to help!
AkashdeepDeb
  • AkashdeepDeb
r = 1/3 So 1/3^n . 3 = 1/6561 1/3^n = 1/19863 n = 9 Now find the sum.
phi
  • phi
you can use wolfram http://www.wolframalpha.com/input/?i=sum+4*%281%2F3%29%5Ei++for+i%3D0+to+8 it found 39364/6561
phi
  • phi
my calculator can change the decimal into that ugly fraction
anonymous
  • anonymous
@phi thank you so much! :) My calculator wouldn't change anything in this problem into fractions.. its a TI=84.. but thanks anyway :)
anonymous
  • anonymous
@AkashdeepDeb thank you too! :)
AkashdeepDeb
  • AkashdeepDeb
:)

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