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

To solve the equation \( \sum_{p=0}^{k}\left(\frac{1}{3} p+\frac{1}{6}\right)=20 \frac{1}{6} \), we first simplify the sum. The expression inside the summation can be separated: \[ \sum_{p=0}^{k}\left(\frac{1}{3} p+\frac{1}{6}\right) = \sum_{p=0}^{k} \frac{1}{3} p + \sum_{p=0}^{k} \frac{1}{6} \] The second term is a constant term summed over \( k+1 \) terms (from \( p=0 \) to \( p=k \)): \[ \sum_{p=0}^{k} \frac{1}{6} = \frac{1}{6} (k + 1) \] Next, we compute the first term, which is a sum of an arithmetic series. The sum of the first \( k \) integers is given by \( \sum_{p=0}^{k} p = \frac{k(k+1)}{2} \): \[ \sum_{p=0}^{k} \frac{1}{3} p = \frac{1}{3} \cdot \frac{k(k+1)}{2} = \frac{k(k+1)}{6} \] Putting these together, we have: \[ \sum_{p=0}^{k}\left(\frac{1}{3} p+\frac{1}{6}\right) = \frac{k(k+1)}{6} + \frac{1}{6}(k + 1) \] Factor out \( \frac{1}{6} \): \[ = \frac{1}{6} [k(k+1) + (k + 1)] = \frac{1}{6} [(k^2 + k + k + 1)] = \frac{1}{6} (k^2 + 2k + 1) = \frac{1}{6} (k + 1)^2 \] Now we set this equal to \( 20 \frac{1}{6} \): \[ \frac{1}{6}(k + 1)^2 = 20 \frac{1}{6} \] Multiplying both sides by \( 6 \): \[ (k + 1)^2 = 20 \] Taking the square root of both sides: \[ k + 1 = \sqrt{20} \text{ or } k + 1 = -\sqrt{20} \] Since we want a non-negative \( k \), we take: \[ k + 1 = \sqrt{20} \] \(\sqrt{20} = 2\sqrt{5}\) which we can also express as approximately \( 4.472 \), but since \( k \) must be a non-negative integer we need to take the closest integer which makes sense: \[ k = 4\sqrt{5} - 1 \text{ must round down to integers as } k = 4 \text{ (since it is the largest integer that satisfies non-negativity)}. \] Thus, the integer \( k = 4 \) is close to \( 4 \) and must satisfy the equation, within non-negative integer constraints: \[ \boxed{4} \]