4) \( \sum_{i=1}^{10}\left(2 i^{3}+5 i\right) \)

To tackle the summation \( \sum_{i=1}^{10}\left(2 i^{3}+5 i\right) \), we can break it into two separate summations: \[ \sum_{i=1}^{10}\left(2 i^{3}+5 i\right) = \sum_{i=1}^{10} 2 i^{3} + \sum_{i=1}^{10} 5 i \] Calculating the first part, \[ \sum_{i=1}^{10} 2 i^{3} = 2 \sum_{i=1}^{10} i^{3} \] The formula for the sum of cubes is \( \left(\frac{n(n+1)}{2}\right)^2 \). For \( n = 10 \): \[ \sum_{i=1}^{10} i^{3} = \left(\frac{10 \cdot 11}{2}\right)^2 = 55^2 = 3025 \] Thus, \[ 2 \sum_{i=1}^{10} i^{3} = 2 \cdot 3025 = 6050 \] Now for the second part, \[ \sum_{i=1}^{10} 5 i = 5 \sum_{i=1}^{10} i \] The formula for the sum of the first \( n \) integers is \( \frac{n(n+1)}{2} \). For \( n = 10 \): \[ \sum_{i=1}^{10} i = \frac{10 \cdot 11}{2} = 55 \] So, \[ 5 \sum_{i=1}^{10} i = 5 \cdot 55 = 275 \] Putting it all together: \[ \sum_{i=1}^{10}\left(2 i^{3}+5 i\right) = 6050 + 275 = 6325 \] Final answer: \[ \sum_{i=1}^{10}\left(2 i^{3}+5 i\right) = 6325 \]