Calculate \( \boldsymbol{n} \) if : \[ \sum_{n=2}(2 n+1)>192 \]

To solve the inequality \( \sum_{n=2}^{m} (2n + 1) > 192 \), we first need to simplify the expression for the sum. The sum can be rewritten as: \[ \sum_{n=2}^{m} (2n + 1) = \sum_{n=2}^{m} 2n + \sum_{n=2}^{m} 1 \] The second sum counts the number of terms from \( n = 2 \) to \( n = m \), which is \( m - 1 \) terms. Thus, we have: \[ \sum_{n=2}^{m} 1 = m - 1 \] Now, we can calculate the first part of the sum: \[ \sum_{n=2}^{m} 2n = 2 \sum_{n=2}^{m} n = 2 \left(\frac{m(m + 1)}{2} - 1\right) \] This simplifies to: \[ = m(m + 1) - 2 \] Therefore, the total sum becomes: \[ \sum_{n=2}^{m} (2n + 1) = m(m + 1) - 2 + (m - 1) = m(m + 1) + m - 3 = m^2 + 2m - 3 \] Now we set up the inequality based on the given problem: \[ m^2 + 2m - 3 > 192 \] Rearranging gives: \[ m^2 + 2m - 195 > 0 \] Next, we'll solve the quadratic equation \( m^2 + 2m - 195 = 0 \) using the quadratic formula: \[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-195)}}{2 \cdot 1} \] \[ = \frac{-2 \pm \sqrt{4 + 780}}{2} = \frac{-2 \pm \sqrt{784}}{2} = \frac{-2 \pm 28}{2} \] Calculating the two potential values for \( m \): 1. \( m = \frac{26}{2} = 13 \) 2. \( m = \frac{-30}{2} = -15 \) (not applicable as \( m \) must be positive) Now we check the inequality for \( m = 13 \): \[ 13^2 + 2 \cdot 13 - 195 = 169 + 26 - 195 = 0 \] To satisfy the inequality \( m^2 + 2m - 195 > 0 \), \( m \) must be greater than 13. Therefore, the smallest integer \( n \) that satisfies the condition is: \[ n = 14 \] Thus, the final answer is: \[ \boxed{14} \]