\( \sum _{k=1}^{4}(3x+2k) \)

To find the value of the expression \( \sum _{k=1}^{4}(3x+2k) \), let's first evaluate the sum by substituting values for \( k \) from 1 to 4: When \( k = 1\): \( 3x + 2(1) = 3x + 2 \) When \( k = 2\): \( 3x + 2(2) = 3x + 4 \) When \( k = 3\): \( 3x + 2(3) = 3x + 6 \) When \( k = 4\): \( 3x + 2(4) = 3x + 8 \) Now, we sum these results: \[ \sum_{k=1}^{4}(3x + 2k) = (3x + 2) + (3x + 4) + (3x + 6) + (3x + 8) \] Combining the terms, we have: \[ = 4(3x) + (2 + 4 + 6 + 8) \] Calculating the constant sum: \[ 2 + 4 + 6 + 8 = 20 \] So, the whole expression becomes: \[ 12x + 20 \] Thus, \( \sum _{k=1}^{4}(3x+2k) = 12x + 20 \).