\( \sum _ { i = 4 } ^ { m } 8 ( \frac { 2 } { 3 } ) ^ { i - 5 } = \frac { 2660 } { 81 } \)

To solve the series, notice that this is a geometric series. The first term when \( i = 4 \) is \( 8 \left( \frac{2}{3} \right)^{-1} = 12 \) and the common ratio is \( \frac{2}{3} \). The number of terms is \( m - 3 \). You can use the formula of the sum of a geometric series, \( S_n = a \frac{1 - r^n}{1 - r} \), where: - \( a \) is the first term, - \( r \) is the common ratio, - \( n \) is the number of terms. In this case, we have: \[ a = 12, r = \frac{2}{3}, n = m - 3. \] Setting up the equation: \[ 12 \frac{1 - \left( \frac{2}{3} \right)^{m-3}}{1 - \frac{2}{3}} = \frac{2660}{81}. \] This simplifies to: \[ 12 \cdot 3 \left( 1 - \left( \frac{2}{3} \right)^{m-3} \right) = \frac{2660}{81}. \] \[ 36 \left( 1 - \left( \frac{2}{3} \right)^{m-3} \right) = \frac{2660}{81}. \] Now, multiply both sides by \( 81 \): \[ 2916 \left( 1 - \left( \frac{2}{3} \right)^{m-3} \right) = 2660. \] This leads to: \[ 1 - \left( \frac{2}{3} \right)^{m-3} = \frac{2660}{2916}. \] Evaluating \( \frac{2660}{2916} \) simplifies to \( \frac{665}{729} \). So, solving for \( \left( \frac{2}{3} \right)^{m-3} \): \[ \left( \frac{2}{3} \right)^{m-3} = 1 - \frac{665}{729} = \frac{64}{729}. \] To solve for \( m \), take the logarithm: \[ m - 3 = \log_{\frac{2}{3}} \left(\frac{2}{3}\right)^6 = 6. \] Thus, \( m = 9 \). Finally, your answer is \( m = 9 \).