\( \sum _ { i = 1 } ^ { n } \frac { i ^ { 2 } + 2 i } { n } \)

To simplify the given summation \( \sum _ { i = 1 } ^ { n } \frac { i ^ { 2 } + 2 i } { n } \), we can first separate the terms inside the summation: \[ \sum _ { i = 1 } ^ { n } \frac { i ^ { 2 } + 2 i } { n } = \frac{1}{n} \sum _ { i = 1 } ^ { n } (i^2 + 2i) \] This then becomes: \[ = \frac{1}{n} \left( \sum _ { i = 1 } ^ { n } i^2 + \sum _ { i = 1 } ^ { n } 2i \right) \] We know the formulas for these summations: 1. The sum of squares: \[ \sum _ { i = 1 } ^ { n } i^2 = \frac{n(n + 1)(2n + 1)}{6} \] 2. The sum of the first \( n \) integers: \[ \sum _ { i = 1 } ^ { n } i = \frac{n(n + 1)}{2} \] Plugging these formulas back into our equation: \[ = \frac{1}{n} \left( \frac{n(n + 1)(2n + 1)}{6} + 2 \cdot \frac{n(n + 1)}{2} \right) \] This simplifies to: \[ = \frac{1}{n} \left( \frac{n(n + 1)(2n + 1)}{6} + n(n + 1) \right) \] Factoring out \( n(n + 1) \): \[ = \frac{1}{n} \left( n(n + 1) \left( \frac{2n + 1}{6} + 1 \right) \right) \] Now simplifying the expression inside the parentheses: \[ \frac{2n + 1}{6} + 1 = \frac{2n + 1 + 6}{6} = \frac{2n + 7}{6} \] Putting it all together: \[ = \frac{1}{n} \left( n(n + 1) \cdot \frac{2n + 7}{6} \right) = \frac{(n + 1)(2n + 7)}{6} \] Thus, the final result for the summation \( \sum _ { i = 1 } ^ { n } \frac { i ^ { 2 } + 2 i } { n } \) is: \[ \frac{(n + 1)(2n + 7)}{6} \]