4. Calculate; \( \sum_{k=1}^{30}(3 k+5) \) 5. Evaluate: \( \sum_{t=0}^{99}(3 t-1) \)

Let's evaluate each of the given summations step by step. --- ### **4. Calculate: \( \sum_{k=1}^{30}(3k + 5) \)** **Step 1: Break Down the Summation** The summation can be separated into two parts: \[ \sum_{k=1}^{30}(3k + 5) = 3 \sum_{k=1}^{30}k + \sum_{k=1}^{30}5 \] **Step 2: Compute Each Part Individually** 1. **Sum of \( k \) from 1 to 30:** \[ \sum_{k=1}^{30}k = \frac{n(n + 1)}{2} = \frac{30 \times 31}{2} = 465 \] 2. **Sum of constants (5) over 30 terms:** \[ \sum_{k=1}^{30}5 = 5 \times 30 = 150 \] **Step 3: Combine the Results** \[ 3 \sum_{k=1}^{30}k + \sum_{k=1}^{30}5 = 3 \times 465 + 150 = 1395 + 150 = 1545 \] **Final Answer:** \[ \sum_{k=1}^{30}(3k + 5) = **1,\!545** \] --- ### **5. Evaluate: \( \sum_{t=0}^{99}(3t - 1) \)** **Step 1: Break Down the Summation** The summation can be separated into two parts: \[ \sum_{t=0}^{99}(3t - 1) = 3 \sum_{t=0}^{99}t - \sum_{t=0}^{99}1 \] **Step 2: Compute Each Part Individually** 1. **Sum of \( t \) from 0 to 99:** \[ \sum_{t=0}^{99}t = \frac{n(n + 1)}{2} = \frac{99 \times 100}{2} = 4,\!950 \] 2. **Sum of constants (-1) over 100 terms (since \( t \) goes from 0 to 99):** \[ \sum_{t=0}^{99}1 = 1 \times 100 = 100 \] **Step 3: Combine the Results** \[ 3 \sum_{t=0}^{99}t - \sum_{t=0}^{99}1 = 3 \times 4,\!950 - 100 = 14,\!850 - 100 = 14,\!750 \] **Final Answer:** \[ \sum_{t=0}^{99}(3t - 1) = **14,\!750** \] --- If you have any further questions or need additional explanations, feel free to ask!