Evaluate the following: (1) \( \sum_{k=1}^{50}(3 k+1) \)

To evaluate the sum \( \sum_{k=1}^{50}(3k+1) \), we can split the summation into two separate components: \[ \sum_{k=1}^{50}(3k+1) = \sum_{k=1}^{50}3k + \sum_{k=1}^{50}1 \] Now we can calculate each part. 1. For the first part \( \sum_{k=1}^{50}3k \): \[ \sum_{k=1}^{50}3k = 3 \sum_{k=1}^{50}k \] The formula for the sum of the first \( n \) natural numbers is: \[ \sum_{k=1}^{n}k = \frac{n(n + 1)}{2} \] Substituting \( n = 50 \): \[ \sum_{k=1}^{50}k = \frac{50(50 + 1)}{2} = \frac{50 \times 51}{2} = 1275 \] Therefore, \[ \sum_{k=1}^{50}3k = 3 \times 1275 = 3825 \] 2. For the second part \( \sum_{k=1}^{50}1 \): This is simply counting the number of terms, which is 50: \[ \sum_{k=1}^{50}1 = 50 \] Now we can combine both parts: \[ \sum_{k=1}^{50}(3k+1) = 3825 + 50 = 3875 \] Thus, the final value of the summation is: \[ \boxed{3875} \]