\( \sum _ { n = 1 } ^ { \infty } ( \frac { 3 } { 5 } ) ^ { n } + \sum _ { n = 2 } ^ { 20 } ( \frac { 3 } { 5 } ) ^ { n } \)

Let’s break down the expression \( \sum _ { n = 1 } ^ { \infty } ( \frac { 3 } { 5 } ) ^ { n } + \sum _ { n = 2 } ^ { 20 } ( \frac { 3 } { 5 } ) ^ { n } \). The first part is an infinite geometric series where the first term is \( a = \frac{3}{5} \) and the common ratio \( r = \frac{3}{5} \). The formula for the sum of an infinite geometric series is: \[ S = \frac{a}{1 - r} \] In this case, \[ S = \frac{\frac{3}{5}}{1 - \frac{3}{5}} = \frac{\frac{3}{5}}{\frac{2}{5}} = \frac{3}{2} \] Now, for the second part \( \sum _ { n = 2 } ^ { 20 } ( \frac { 3 } { 5 } ) ^ { n } \), we rewrite it as: \[ \sum_{n=2}^{20} \left(\frac{3}{5}\right)^n = \left(\frac{3}{5}\right)^2 + \left(\frac{3}{5}\right)^3 + \cdots + \left(\frac{3}{5}\right)^{20} \] This is a finite geometric series where the first term \( a = \left(\frac{3}{5}\right)^2 \) and the common ratio \( r = \frac{3}{5} \). The number of terms \( n \) is from 2 to 20, so there are \( 20 - 2 + 1 = 19 \) terms. The formula for the sum of a finite geometric series is: \[ S_n = a \frac{1 - r^n}{1 - r} \] Thus, we have: \[ S_{19} = \left(\frac{3}{5}\right)^2 \frac{1 - \left(\frac{3}{5}\right)^{19}}{1 - \frac{3}{5}} = \left(\frac{3}{5}\right)^2 \cdot \frac{1 - \left(\frac{3}{5}\right)^{19}}{\frac{2}{5}} = \frac{9}{25} \cdot \frac{5}{2} \cdot \left( 1 - \left(\frac{3}{5}\right)^{19} \right) = \frac{9}{10} \left( 1 - \left(\frac{3}{5}\right)^{19} \right) \] Now we can sum both parts together: \[ \sum _ { n = 1 } ^ { \infty } \left( \frac { 3 } { 5 } \right) ^ { n } + \sum _ { n = 2 } ^ { 20 } \left( \frac { 3 } { 5 } \right) ^ { n } = \frac{3}{2} + \frac{9}{10} \left( 1 - \left(\frac{3}{5}\right)^{19} \right) \] Finally, combining these two sums gives us the final result. **So the final answer is:** \[ \frac{3}{2} + \frac{9}{10} \left( 1 - \left(\frac{3}{5}\right)^{19} \right) \]