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

benjamingu22  May 27, 2017
 #1
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The sum of this inifnite series  =

 

4 / ( 1 - 3/a)        where  4  is the first term  and  (3/a)  is the common ratio

 

Simplifying this, we have

 

4a /  [ a - 3]

 

 

Note.........that  since a is positive, the first value of a that produces a perfect square is when a  = 4

 

4(4) / [ 4 - 3 ]  =  16 / 1    =  16    which is a perfect square

 

 

 

cool cool cool

CPhill  May 27, 2017

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