4i. If \( \sum_{i=1}^{n}(i-2)=252 \), determine the number of terms in the series.

To solve the equation \( \sum_{i=1}^{n}(i-2)=252 \), we first rewrite the summation. The sum can be expressed as: \[ \sum_{i=1}^{n}(i-2) = \sum_{i=1}^{n}i - \sum_{i=1}^{n}2 \] The first part, \( \sum_{i=1}^{n}i \), is the well-known formula for the sum of the first \( n \) integers: \[ \sum_{i=1}^{n}i = \frac{n(n+1)}{2} \] The second part, \( \sum_{i=1}^{n}2 \), is simply: \[ \sum_{i=1}^{n}2 = 2n \] Putting this all together, we have: \[ \sum_{i=1}^{n}(i-2) = \frac{n(n+1)}{2} - 2n = \frac{n(n+1) - 4n}{2} = \frac{n^2 - 3n}{2} \] Setting this equal to 252 gives us: \[ \frac{n^2 - 3n}{2} = 252 \] Multiplying both sides by 2 to eliminate the fraction: \[ n^2 - 3n = 504 \] Rearranging this into a standard quadratic equation form: \[ n^2 - 3n - 504 = 0 \] To solve this quadratic equation, we can use the quadratic formula: \[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = -3 \), and \( c = -504 \). Plugging these values into the formula: \[ n = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-504)}}{2 \cdot 1} \] \[ = \frac{3 \pm \sqrt{9 + 2016}}{2} \] \[ = \frac{3 \pm \sqrt{2025}}{2} \] Calculating the square root: \[ \sqrt{2025} = 45 \] Now substituting back: \[ n = \frac{3 \pm 45}{2} \] This gives us two potential solutions: 1. \( n = \frac{48}{2} = 24 \) 2. \( n = \frac{-42}{2} = -21 \) (not a valid solution since \( n \) must be positive) Thus, the number of terms in the series is: \[ \boxed{24} \]