If \( \sum_{p=0}^{x}\left(\frac{1}{3} p+\frac{1}{6}\right)=20 \frac{1}{6} \), determine the value of \( k \)

To solve the equation \[ \sum_{p=0}^{x}\left(\frac{1}{3} p+\frac{1}{6}\right)=20 \frac{1}{6}, \] we start by rewriting \( 20 \frac{1}{6} \) as an improper fraction: \[ 20 \frac{1}{6} = \frac{120 + 1}{6} = \frac{121}{6}. \] Next, we can simplify the summation: \[ \sum_{p=0}^{x} \left(\frac{1}{3} p + \frac{1}{6}\right) = \sum_{p=0}^{x} \frac{1}{3} p + \sum_{p=0}^{x} \frac{1}{6}. \] The second summation can be calculated easily: \[ \sum_{p=0}^{x} \frac{1}{6} = (x + 1) \cdot \frac{1}{6} = \frac{x + 1}{6}. \] For the first summation, we utilize the formula for the sum of the first \( n \) integers: \[ \sum_{p=0}^{x} p = \frac{x(x + 1)}{2}. \] Thus, \[ \sum_{p=0}^{x} \frac{1}{3} p = \frac{1}{3} \cdot \frac{x(x + 1)}{2} = \frac{x(x + 1)}{6}. \] Now, we combine both parts of the summation: \[ \sum_{p=0}^{x}\left(\frac{1}{3} p + \frac{1}{6}\right) = \frac{x(x + 1)}{6} + \frac{x + 1}{6} = \frac{x(x + 1) + (x + 1)}{6} = \frac{x(x + 1) + (x + 1)}{6} = \frac{(x + 1)(x + 2)}{6}. \] We can now set this equal to \( \frac{121}{6} \): \[ \frac{(x + 1)(x + 2)}{6} = \frac{121}{6}. \] Multiplying both sides by \( 6 \) to eliminate the denominator yields: \[ (x + 1)(x + 2) = 121. \] Next, expanding the left-hand side gives: \[ x^2 + 3x + 2 = 121. \] Rearranging to bring all terms to one side results in: \[ x^2 + 3x + 2 - 121 = 0 \quad \Rightarrow \quad x^2 + 3x - 119 = 0. \] Now, we will solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-119)}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 + 476}}{2} = \frac{-3 \pm \sqrt{485}}{2}. \] We need to simplify \( \sqrt{485} \): \[ 485 = 5 \times 97 \quad \Rightarrow \quad \sqrt{485} = \sqrt{5} \sqrt{97}. \] Thus, \[ x = \frac{-3 \pm \sqrt{485}}{2}. \] Since \( x \) must be a non-negative integer, we take the positive root: \[ x = \frac{-3 + \sqrt{485}}{2}. \] Calculating \( \sqrt{485} \) yields approximately 22.036 (or slightly more than 22), giving us \[ x \approx \frac{-3 + 22}{2} \approx \frac{19}{2} \approx 9.5. \] As \( x \) must be an integer, we round it to the nearest integers: For \( x = 9 \): \[ (9 + 1)(9 + 2) = 10 \cdot 11 = 110 \quad (not \, correct) \] For \( x = 10 \): \[ (10 + 1)(10 + 2) = 11 \cdot 12 = 132 \quad (not \, correct) \] For a systematic trial checking, we