\( \sum _ { t = - 2 } ^ { 30 } ( 4 t + 6 ) \)

To solve the sum \( \sum _ { t = - 2 } ^ { 30 } ( 4 t + 6 ) \), we can first simplify the expression being summed. The sum can be rewritten as: \[ \sum _ { t = - 2 } ^ { 30 } ( 4 t + 6 ) = \sum _ { t = - 2 } ^ { 30 } 4t + \sum _ { t = - 2 } ^ { 30 } 6 \] Calculating the first part: \[ \sum _ { t = - 2 } ^ { 30 } 4t = 4 \sum _ { t = - 2 } ^ { 30 } t \] Now, we can compute the sum \( \sum _ { t = - 2 } ^ { 30 } t \). The range of \( t \) is from -2 to 30, inclusive. The number of terms in this sum can be found as follows: \[ n = 30 - (-2) + 1 = 30 + 2 + 1 = 33 \] The sum of an arithmetic sequence can be calculated using the formula: \[ \text{Sum} = \frac{n}{2} \times (\text{first term} + \text{last term}) \] Here, \( n = 33 \), the first term is -2, and the last term is 30: \[ \sum _ { t = - 2 } ^ { 30 } t = \frac{33}{2} \times (-2 + 30) = \frac{33}{2} \times 28 = 33 \times 14 = 462 \] Now substituting this back into our earlier equation: \[ \sum _ { t = - 2 } ^ { 30 } 4t = 4 \times 462 = 1848 \] Next, we calculate the second part of the sum: \[ \sum _ { t = - 2 } ^ { 30 } 6 = 6 \times 33 = 198 \] Now we combine both parts together: \[ \sum _ { t = - 2 } ^ { 30 } ( 4t + 6 ) = 1848 + 198 = 2046 \] Thus, the final result is: \[ \boxed{2046} \]