Final answer:
The question revolves around a geometric series where the sum is a perfect square. The answer involves finding the smallest possible positive integer value for n such that 2n^2 is also a perfect square, which in this case is n = 2.
Step-by-step explanation:
The question appears to be related to geometric series and perfect squares. Although there are typographical errors in the original query, the essence of the question seems to be about finding the smallest possible value for a variable, given that the sum of a geometric series is a perfect square, and where the variable in question is a positive integer.
Let's consider a geometric series with n terms, where the sum S is a perfect square. The sum of the first n terms of an arithmetic series of odd numbers can be represented as S = n2. By manipulating the series, for example, by adding and subtracting terms within the series, we get 2[1 + (n - 1) + 3 + ... + (2n - 3) + (2n - 1) - (n - 1)] which simplifies to 2[n2].
If we want this expression to be a perfect square, we need to deduce that the smallest possible value for n will be the case where 2n2 = m2 for some integer m. We can see that n must be a multiple of the square root of 2, and the smallest such positive integer is 1. Therefore, 2(1)2 = 2, which happens to be the perfect square of √2. However, since √2 is not an integer, we seek the next smallest integer value where 2n2 is a perfect square, which is n = √2√2 = 2. Thus the smallest possible integer value for n for which 2n2 is a perfect square is 2.