Evaluate \( \sum_{i=1}^{8} 5 \times 3^{i-1} \) Give an exact answer, do not round.

To evaluate the sum \( \sum_{i=1}^{8} 5 \times 3^{i-1} \), we can factor out the constant \( 5 \) since it does not depend on \( i \): \[ \sum_{i=1}^{8} 5 \times 3^{i-1} = 5 \sum_{i=1}^{8} 3^{i-1} \] Next, we need to evaluate the sum \( \sum_{i=1}^{8} 3^{i-1} \). This is a geometric series where the first term \( a = 3^0 = 1 \) and the common ratio \( r = 3 \). The number of terms \( n = 8 \). The formula for the sum \( S_n \) of the first \( n \) terms of a geometric series is given by: \[ S_n = a \frac{1 - r^n}{1 - r} \] Substituting \( a = 1 \), \( r = 3 \), and \( n = 8 \) into the formula, we get: \[ S_8 = 1 \frac{1 - 3^8}{1 - 3} = \frac{1 - 3^8}{-2} \] Calculating \( 3^8 \): \[ 3^8 = 6561 \] Now substituting this value back into our expression for \( S_8 \): \[ S_8 = \frac{1 - 6561}{-2} = \frac{-6560}{-2} = 3280 \] Now we can substitute \( S_8 \) back into our first equation: \[ \sum_{i=1}^{8} 5 \times 3^{i-1} = 5 \times 3280 = 16400 \] Thus, the exact answer for \( \sum_{i=1}^{8} 5 \times 3^{i-1} \) is: \[ \boxed{16400} \]