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Evaluate the infinite geometric series: \(\frac{3}{2}-\frac{2}{3}+\frac{8}{27}-\frac{32}{243}+\dots\)

 Dec 27, 2018

Best Answer 

 #2
avatar+3715 
+2

We can also use our handy formula: \(\frac{n}{1-r}.\) Since the series has a first term of \(\frac{3}{2}\) and a common ratio of \(\frac{-4}{9}\), we attain \(\cfrac{\frac{3}{2}}{1-\left(\frac{-4}{9}\right)}=\boxed{\frac{27}{26}}.\)

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 Dec 27, 2018
 #1
avatar+94209 
+2

Notice that we can split this up as

 

3/2  + 8/27 +.......     and

 

-2/3  - 32/243 +   .......  =  - ( 2/3 + 32/243 + .....)

 

The geometric ratio of the  first series is  (2/3)^4  =  16/81

 

So.....the sum of this series  =   (3/2)  / ( 1 - 16/81)  = (3/2) / ( 65/81) =   243/130

 

The geometric ratio of the terms in the parentheses in the second series is  ( 2/3)^4 = 16/81

 

So  the sum of the second series is

 

-  [ (2/3) / ( 1 - 16/81 ) ]  =   -  ( 2/3) / ( 65/81)    =   - (2/3) (81/65) =  - 54 /65 = -108 / 130

 

 

So....the sum is   [243 - 108] / 130  =  135 / 130   =  27 /26

 

 

cool cool cool

 Dec 27, 2018
edited by CPhill  Dec 27, 2018
 #2
avatar+3715 
+2
Best Answer

We can also use our handy formula: \(\frac{n}{1-r}.\) Since the series has a first term of \(\frac{3}{2}\) and a common ratio of \(\frac{-4}{9}\), we attain \(\cfrac{\frac{3}{2}}{1-\left(\frac{-4}{9}\right)}=\boxed{\frac{27}{26}}.\)

tertre Dec 27, 2018
 #3
avatar+94209 
+1

Thanks, tertre.....your method is better than mine...I even forgot that my final fraction could be simplified even more!!!

 

 

Good job  !!!

 

 

 

cool cool cool

CPhill  Dec 27, 2018
 #4
avatar+3715 
+2

Thank you, CPhill! I just learned this a few days ago... laugh

tertre  Dec 27, 2018
 #6
avatar+534 
+1

Wow tertre, that's REALLY cool! :D

CoolStuffYT  Dec 29, 2018
 #5
avatar+836 
+1

Thank you, everyone! 

 Dec 28, 2018

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