Questions: Question 17 (1 point) Evaluate the sum from i=1 to n of (i^2 + 1) = - 5n^3 + 7n / 6 - 2n^3 + 3n^2 + 7 / 6 - 2n^3 + 3n^2 + 7n / 6 - 3n^3 + 2n^2 + 7n / 6 - Infinity

Evaluate the sum \( \sum_{i=1}^{n}\left(i^{2}+1\right) \) for \( n = 5 \).

Calculate the sum of squares.

Using the formula \( \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \), we find: \[ \sum_{i=1}^{5} i^2 = \frac{5(5+1)(2 \cdot 5 + 1)}{6} = \frac{5 \cdot 6 \cdot 11}{6} = 55. \]

Calculate the sum of ones.

The sum of 1 from 1 to \( n \) is simply \( n \): \[ \sum_{i=1}^{5} 1 = 5. \]

Combine the results.

Adding the two results gives: \[ \sum_{i=1}^{5}\left(i^{2}+1\right) = 55 + 5 = 60. \]

The answer is \( \boxed{60} \).

The final answer is \( \boxed{60} \).