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# Geometric Series

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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

#2
+4614
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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 $$\cfrac{\frac{3}{2}}{1-\left(\frac{-4}{9}\right)}=\boxed{\frac{27}{26}}.$$

Dec 27, 2018

#1
+128845
+3

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

Dec 27, 2018
edited by CPhill  Dec 27, 2018
#2
+4614
+3

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
+128845
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Thanks, tertre.....your method is better than mine...I even forgot that my final fraction could be simplified even more!!!

Good job  !!!

CPhill  Dec 27, 2018
#4
+4614
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Thank you, CPhill! I just learned this a few days ago...

tertre  Dec 27, 2018
#6
+1252
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Wow tertre, that's REALLY cool! :D

CoolStuffYT  Dec 29, 2018
#5
+884
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Thank you, everyone!

Dec 28, 2018