+0  
 
+4
263
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avatar+738 

Consider the geometric series \(4+\frac{12}{a}+\frac{36}{a^2}+\cdots.\) If the sum is a perfect square, what is the smallest possible value of a where a is a positive integer?

MIRB16  Sep 11, 2017
 #1
avatar+91360 
+2

 

Note that 

 

r = 3/a  

 

And the sum of the series can be represented as

 

S = 4 / [ 1 - 3/a ]  =  4a / [ a - 3]

 

Note that if a = 4, then the sum  is   4 (4) / [ 4 - 3]  =  16/ 1  =  16   which is a perfect square 

 

 

 

cool cool cool

CPhill  Sep 11, 2017

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