Find the sum of the geometric series, if possible. \[ 3+9+27+81+\ldots \]

To find the sum of the geometric series \( 3 + 9 + 27 + 81 + \ldots \), we first identify the first term and the common ratio. The first term \( a \) is: \[ a = 3 \] Next, we find the common ratio \( r \). We can calculate \( r \) by dividing the second term by the first term: \[ r = \frac{9}{3} = 3 \] Now, we can express the series in a more general form. The series can be written as: \[ S = 3 + 3 \cdot 3 + 3 \cdot 3^2 + 3 \cdot 3^3 + \ldots \] This can be rewritten as: \[ S = 3(1 + 3 + 3^2 + 3^3 + \ldots) \] The series inside the parentheses is an infinite geometric series with first term \( 1 \) and common ratio \( 3 \). The formula for the sum of an infinite geometric series is given by: \[ S = \frac{a}{1 - r} \] where \( |r| < 1 \). In our case, the common ratio \( r = 3 \), which does not satisfy the condition \( |r| < 1 \). Therefore, the series diverges. Since the series diverges, we conclude that the sum of the series \( 3 + 9 + 27 + 81 + \ldots \) is: \[ \text{The series diverges.} \]