Evaluate \[ \sum_{k=1}^{21} 5 k-12 \] \( \sum_{n=1}^{21}(6 n+3) \)

To evaluate the expression \( \sum_{k=1}^{21} (5k - 12) \): First, we can break the sum into two separate sums: \[ \sum_{k=1}^{21} (5k - 12) = \sum_{k=1}^{21} 5k - \sum_{k=1}^{21} 12 \] Calculating each part: 1. The first sum \( \sum_{k=1}^{21} 5k \) can be factored out: \[ \sum_{k=1}^{21} 5k = 5 \sum_{k=1}^{21} k = 5 \cdot \frac{21 \cdot 22}{2} = 5 \cdot 231 = 1155 \] 2. The second sum \( \sum_{k=1}^{21} 12 \) can be calculated as: \[ \sum_{k=1}^{21} 12 = 12 \cdot 21 = 252 \] Putting it all together: \[ \sum_{k=1}^{21} (5k - 12) = 1155 - 252 = 903 \] Next, we evaluate the second expression \( \sum_{n=1}^{21} (6n + 3) \): Similarly, we can split this sum as well: \[ \sum_{n=1}^{21} (6n + 3) = \sum_{n=1}^{21} 6n + \sum_{n=1}^{21} 3 \] Calculating each part: 1. The first sum \( \sum_{n=1}^{21} 6n \): \[ \sum_{n=1}^{21} 6n = 6 \sum_{n=1}^{21} n = 6 \cdot \frac{21 \cdot 22}{2} = 6 \cdot 231 = 1386 \] 2. The second sum \( \sum_{n=1}^{21} 3 \): \[ \sum_{n=1}^{21} 3 = 3 \cdot 21 = 63 \] Now, adding these together: \[ \sum_{n=1}^{21} (6n + 3) = 1386 + 63 = 1449 \] Thus, the final results are: \[ \sum_{k=1}^{21} (5k - 12) = 903 \] \[ \sum_{n=1}^{21} (6n + 3) = 1449 \]