\( \sum_{t=-2}^{30}(4 t+6) \) Evaluate

To evaluate the sum \( \sum_{t=-2}^{30}(4 t+6) \), we can start by simplifying the expression inside the summation: \[ 4t + 6 \] First, we will find the number of terms in the sum. The sum runs from \( t = -2 \) to \( t = 30 \). The number of terms is: \[ 30 - (-2) + 1 = 30 + 2 + 1 = 33 \] Next, we can express the sum of the linear expression. Decomposing the sum: \[ \sum_{t=-2}^{30} (4t + 6) = \sum_{t=-2}^{30} 4t + \sum_{t=-2}^{30} 6 \] The second sum is straightforward since there are 33 terms, each contributing 6: \[ \sum_{t=-2}^{30} 6 = 6 \times 33 = 198 \] Now we calculate the first sum \( \sum_{t=-2}^{30} 4t \): \[ \sum_{t=-2}^{30} 4t = 4 \sum_{t=-2}^{30} t \] To find \( \sum_{t=-2}^{30} t \), we use the formula for the sum of an arithmetic series: \[ \sum_{k=a}^{b} k = \frac{n}{2} (a+b) \] where \( n \) is the number of terms, \( a \) is the first term, and \( b \) is the last term. Here, \( a = -2 \), \( b = 30 \), and \( n = 33 \). Thus, \[ \sum_{t=-2}^{30} t = \frac{33}{2}((-2) + 30) = \frac{33}{2}(28) = 33 \times 14 = 462 \] Now we can return to the first sum: \[ \sum_{t=-2}^{30} 4t = 4 \cdot 462 = 1848 \] Combining both parts of the sum: \[ \sum_{t=-2}^{30} (4t + 6) = 1848 + 198 = 2046 \] Thus, the final answer is: \[ \boxed{2046} \]