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Let
a + ar + ar^2 + ar^3 + \dotsb
be an infinite geometric series.  The sum of the series is $2.$  The sum of the cubes of all the terms is $4.$ Find the common ratio.

 Mar 31, 2024
 #1
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a / (1 - r)   = 2

 

a  = 2(1 - r)   →    a^3  = 8 ( 1 - r)^3         (1)

 

And

 

a^3  + (ar)^3  + (ar^2)^3 + (ar^3)^3  + ......  =   4

 

a^3  / ( 1 - r^3)  = 4

 

a^3  = 4 ( 1 - r^3)    (2)

 

Equate (1), (2)

 

8 ( 1 - r)^3  = 4 (1 - r^3)

 

8 ( 1 - r)^3  = 4 ( 1 - r) (1 + r + r^2)

 

8 (1 -r)^2  = 4 ( 1 + r + r^2)

 

8 (r^2 - 2r + 1) =  4 ( 1 + r + t^2)

 

2r^2 - 4r + 2  = 1 + r + r^2

 

r^2 - 5r + 1  =  0

 

r  =  [ 5 -  sqrt [ 25 - 4] ] /  2   = [   5 - sqrt (21) ]  / 2

 

 

cool cool cool

 Apr 1, 2024
edited by CPhill  Apr 1, 2024

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