\( \sum _ { k = 1 } ^ { m } ( k + 1 ) = 1595 \)

To solve the equation \( \sum _ { k = 1 } ^ { m } ( k + 1 ) = 1595 \), we first rewrite the sum: \[ \sum _ { k = 1 } ^ { m } ( k + 1 ) = \sum _ { k = 1 } ^ { m } k + \sum _ { k = 1 } ^ { m } 1 = \sum _ { k = 1 } ^ { m } k + m \] Using the formula for the sum of the first \( m \) integers: \[ \sum _ { k = 1 } ^ { m } k = \frac{m(m + 1)}{2} \] So, we can rewrite our equation as: \[ \frac{m(m + 1)}{2} + m = 1595 \] This simplifies to: \[ \frac{m(m + 1) + 2m}{2} = 1595 \] Multiplying both sides by 2, we get: \[ m(m + 1) + 2m = 3190 \] This further simplifies to: \[ m^2 + 3m - 3190 = 0 \] Now we can solve this quadratic equation using the quadratic formula \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): Here, \( a = 1 \), \( b = 3 \), and \( c = -3190 \): \[ m = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-3190)}}{2 \cdot 1} \] Calculating the discriminant: \[ 3^2 + 4 \cdot 3190 = 9 + 12760 = 12769 \] Now, finding the square root of \( 12769 \): \[ \sqrt{12769} = 113 \] Now substituting back into the quadratic formula: \[ m = \frac{-3 \pm 113}{2} \] Calculating the two possible values for \( m \): 1. \( m = \frac{110}{2} = 55 \) 2. \( m = \frac{-116}{2} = -58 \) (not applicable since \( m \) must be positive) Thus, the solution is: \[ m = 55 \] Finally, we can verify this: \[ \sum_{k=1}^{55} (k + 1) = \sum_{k=1}^{55} k + 55 = \frac{55 \times 56}{2} + 55 = 1540 + 55 = 1595 \] Therefore, the value of \( m \) is: \[ \boxed{55} \]