\( 3 ( - \frac { 1 } { 3 } ) + 3 ( - \frac { 1 } { 3 } ) ^ { 2 } + 3 ( - \frac { 1 } { 3 } ) ^ { 3 } + \ldots = \square \)

To solve the series \[ 3 \left( -\frac{1}{3} \right) + 3 \left( -\frac{1}{3} \right)^2 + 3 \left( -\frac{1}{3} \right)^3 + \ldots, \] we can factor out the common term \(3\): \[ 3 \left( -\frac{1}{3} + \left( -\frac{1}{3} \right)^2 + \left( -\frac{1}{3} \right)^3 + \ldots \right). \] Now, we need to analyze the series inside the parentheses: \[ -\frac{1}{3} + \left( -\frac{1}{3} \right)^2 + \left( -\frac{1}{3} \right)^3 + \ldots. \] This is an infinite geometric series where the first term \(a = -\frac{1}{3}\) and the common ratio \(r = -\frac{1}{3}\). The formula for the sum \(S\) of an infinite geometric series is given by: \[ S = \frac{a}{1 - r}, \] provided that \(|r| < 1\). In our case, we have: - \(a = -\frac{1}{3}\) - \(r = -\frac{1}{3}\) Since \(|r| = \frac{1}{3} < 1\), we can apply the formula: \[ S = \frac{-\frac{1}{3}}{1 - \left(-\frac{1}{3}\right)} = \frac{-\frac{1}{3}}{1 + \frac{1}{3}} = \frac{-\frac{1}{3}}{\frac{4}{3}} = -\frac{1}{3} \cdot \frac{3}{4} = -\frac{1}{4}. \] Now substituting this back into our expression, we have: \[ 3 \left( -\frac{1}{4} \right) = -\frac{3}{4}. \] Thus, the value of the original series is \[ \boxed{-\frac{3}{4}}. \]