+98 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?
+128407
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
