\( \frac { 1 } { 3 } + \frac { 1 } { 3 ^ { 2 } } + \ldots + \frac { 1 } { 3 ^ { 2024 } } \)

The given series is \[ S = \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \ldots + \frac{1}{3^{2024}}. \] This is a geometric series where the first term \( a = \frac{1}{3} \) and the common ratio \( r = \frac{1}{3} \). The number of terms in the series is 2024. The formula for the sum of the first \( n \) terms of a geometric series is given by: \[ S_n = a \frac{1 - r^n}{1 - r}, \] where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms. In our case, we have: - \( a = \frac{1}{3} \) - \( r = \frac{1}{3} \) - \( n = 2024 \) Substituting these values into the formula, we get: \[ S = \frac{1}{3} \cdot \frac{1 - \left(\frac{1}{3}\right)^{2024}}{1 - \frac{1}{3}}. \] Calculating \( 1 - \frac{1}{3} \): \[ 1 - \frac{1}{3} = \frac{2}{3}. \] Now substituting this back into the sum formula: \[ S = \frac{1}{3} \cdot \frac{1 - \left(\frac{1}{3}\right)^{2024}}{\frac{2}{3}}. \] This simplifies to: \[ S = \frac{1}{3} \cdot \frac{3}{2} \left(1 - \left(\frac{1}{3}\right)^{2024}\right). \] The \( \frac{3}{3} \) cancels out: \[ S = \frac{1}{2} \left(1 - \left(\frac{1}{3}\right)^{2024}\right). \] Thus, the final result for the sum of the series is: \[ \boxed{\frac{1}{2} \left(1 - \frac{1}{3^{2024}}\right)}. \]