3. Determine the smallest number of terms for which... \( \sum_{r=1}^{k}(3 r-2)>117 \)

To solve the inequality \( \sum_{r=1}^{k}(3r-2) > 117 \), we first need to express the sum more explicitly. The sum \( \sum_{r=1}^{k}(3r-2) \) can be separated into two parts: \[ \sum_{r=1}^{k}(3r-2) = \sum_{r=1}^{k}(3r) - \sum_{r=1}^{k}(2) \] Calculating each component: - The sum \( \sum_{r=1}^{k}(3r) = 3 \sum_{r=1}^{k} r = 3 \cdot \frac{k(k+1)}{2} = \frac{3k(k+1)}{2} \) - The sum \( \sum_{r=1}^{k}(2) = 2k \) Putting these together, we have: \[ \sum_{r=1}^{k}(3r-2) = \frac{3k(k+1)}{2} - 2k \] Simplifying this: \[ \sum_{r=1}^{k}(3r-2) = \frac{3k(k+1)}{2} - \frac{4k}{2} = \frac{3k^2 + 3k - 4k}{2} = \frac{3k^2 - k}{2} = \frac{k(3k - 1)}{2} \] Now we need to solve the inequality: \[ \frac{k(3k - 1)}{2} > 117 \] Multiplying both sides by 2 to eliminate the fraction: \[ k(3k - 1) > 234 \] Rearranging gives us the quadratic inequality: \[ 3k^2 - k - 234 > 0 \] To find the critical points, we can use the quadratic formula: \[ k = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 3 \cdot (-234)}}{2 \cdot 3} = \frac{1 \pm \sqrt{1 + 2808}}{6} = \frac{1 \pm \sqrt{2809}}{6} \] Calculating \( \sqrt{2809} \): \[ \sqrt{2809} = 53 \Rightarrow k = \frac{1 \pm 53}{6} \] This gives us two potential solutions: \[ k = \frac{54}{6} = 9 \quad \text{and} \quad k = \frac{-52}{6} \text{ (discarded since } k \text{ must be positive)} \] Now we need to determine for which values of \( k > 9 \) the inequality holds. Evaluating at \( k = 9 \): \[ 3(9)(9 - 1) = 3(9)(8) = 216 \] And at \( k = 10 \): \[ 3(10)(10 - 1) = 3(10)(9) = 270 \] Since \( 216 > 234 \) is false, and \( 270 > 234 \) is true, the smallest integer \( k \) satisfying the initial inequality is: \[ \boxed{10} \]