What is the greatest number of terms for which the series \( \sum_{k=1}^{n}(k+1) \) will have a value less than 65 ?

To find the greatest number of terms \( n \) for which the series \( \sum_{k=1}^{n} (k+1) \) is less than 65, let's first express the sum. The series can be rewritten as: \[ \sum_{k=1}^{n} (k+1) = \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 = \frac{n(n+1)}{2} + n = \frac{n(n+1)}{2} + \frac{2n}{2} = \frac{n(n+1) + 2n}{2} = \frac{n^2 + 3n}{2} \] Now, we set up the inequality: \[ \frac{n^2 + 3n}{2} < 65 \] Multiplying both sides by 2 to eliminate the fraction gives: \[ n^2 + 3n < 130 \] Rearranging this leads to: \[ n^2 + 3n - 130 < 0 \] Next, we will find the roots of the quadratic equation \( n^2 + 3n - 130 = 0 \) using the quadratic formula: \[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-130)}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 + 520}}{2} = \frac{-3 \pm \sqrt{529}}{2} \] \[ \sqrt{529} = 23 \] Thus, the roots are: \[ n = \frac{-3 + 23}{2} = \frac{20}{2} = 10 \quad \text{and} \quad n = \frac{-3 - 23}{2} = \frac{-26}{2} = -13 \] Since we are looking for positive \( n \), we focus on \( n = 10 \). Now we should check the value of the series at \( n = 10 \) and \( n = 11 \): For \( n = 10 \): \[ \sum_{k=1}^{10} (k+1) = \frac{10^2 + 3 \times 10}{2} = \frac{100 + 30}{2} = \frac{130}{2} = 65 \] For \( n = 11 \): \[ \sum_{k=1}^{11} (k+1) = \frac{11^2 + 3 \times 11}{2} = \frac{121 + 33}{2} = \frac{154}{2} = 77 \] Thus, the greatest number of terms \( n \) for which the series \( \sum_{k=1}^{n} (k+1) \) has a value less than 65 is: \[ \boxed{9} \]