\( \sum _ { i = 1 } ^ { n } ( 3 i - 5 ) = 3430 \)

To find \( n \) for the equation \( \sum _ { i = 1 } ^ { n } ( 3 i - 5 ) = 3430 \), we can simplify the left-hand side. The sum can be split into two parts: \[ \sum _ { i = 1 } ^ { n } ( 3 i - 5 ) = \sum _ { i = 1 } ^ { n } 3i - \sum _ { i = 1 } ^ { n } 5 \] Calculating each part, we have: \[ \sum _ { i = 1 } ^ { n } 3i = 3 \sum _ { i = 1 } ^ { n } i = 3 \cdot \frac{n(n+1)}{2} = \frac{3n(n+1)}{2} \] And \[ \sum _ { i = 1 } ^ { n } 5 = 5n \] Now, putting it all together: \[ \frac{3n(n+1)}{2} - 5n = 3430 \] Multiplying every term by 2 to eliminate the fraction: \[ 3n(n+1) - 10n = 6860 \] This simplifies to: \[ 3n^2 + 3n - 10n - 6860 = 0 \implies 3n^2 - 7n - 6860 = 0 \] Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ n = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 3 \cdot (-6860)}}{2 \cdot 3} \] Calculating the discriminant: \[ b^2 - 4ac = 49 + 4 \cdot 3 \cdot 6860 = 49 + 82320 = 82369 \] \[ \sqrt{82369} = 287 \quad (\text{since } 287^2 = 82369) \] Now putting it back into the formula: \[ n = \frac{7 \pm 287}{6} \] Calculating the two potential values for \( n \): 1. \( n = \frac{294}{6} = 49 \) 2. \( n = \frac{-280}{6} \quad (\text{which is negative and not valid}) \) Thus, the only valid solution is: \[ n = 49 \] The value of \( n \) is \( 49 \).