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In a certain infinite geometric series, the first term is 1, and each term is equal to the sum of the two terms after it. (For example, the first term is equal to the sum of the second term and third term). If $s$ is the sum of the series and $r$ is the common ratio, find $r-s$.

 Mar 2, 2019
 #1
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+1

Solve: r^2 + r - 1
r =1/2[sqrt(5) - 1]
s =2.6180339887
r - s = - 2

 Mar 3, 2019
 #2
avatar+101856 
+3

First term  1 = r + r^2   →  r = 1 - r^2   →   r^2 = 1 - r

 

Sum  =     1 / (1 - r )   =      1  / [ 1 - (1 - r^2) ]      =   1 / r^2    

 

r - s  =

 

r - 1/r^2  =

 

[ r^3 - 1 ] / [  1 - r ] =

 

- [ 1 - r^3 ] / [ 1 - r ] =

 

- [ 1 - r ] [ 1 + r + r^2 ] / [ 1 - r ]  =

 

- [ 1 + r + r^2 ] =

 

- [ 1 + (r + r^2 ) ] =

 

- [ 1 + 1 ] =

 

-2

 

cool cool cool

 Mar 3, 2019
edited by CPhill  Mar 3, 2019
edited by CPhill  Mar 3, 2019

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