Find the sum. \( \sum_{i=1}^{3!}(-3-4 i) \) The sum is T.

To find the sum \( \sum_{i=1}^{3!}(-3-4i) \), we first compute \( 3! = 6 \). So, we need to sum from \( i=1 \) to \( i=6 \). The sum can be expanded as follows: \[ \sum_{i=1}^{6} (-3 - 4i) = \sum_{i=1}^{6} (-3) + \sum_{i=1}^{6} (-4i) \] Calculating the first part: \[ \sum_{i=1}^{6} (-3) = -3 \times 6 = -18 \] Now, for the second part: \[ \sum_{i=1}^{6} (-4i) = -4 \sum_{i=1}^{6} i \] Using the formula for the sum of the first \( n \) integers \( \sum_{i=1}^{n} i = \frac{n(n+1)}{2} \), we find: \[ \sum_{i=1}^{6} i = \frac{6(6 + 1)}{2} = \frac{6 \times 7}{2} = 21 \] Thus, \[ \sum_{i=1}^{6} (-4i) = -4 \times 21 = -84 \] Now, combining both parts: \[ \sum_{i=1}^{6} (-3 - 4i) = -18 - 84 = -102 \] Therefore, the sum \( T \) is: \[ \boxed{-102} \]