3.3 Calculate: \( \sum_{k=3}^{5}(-1)^{k} \cdot \frac{2}{k} \)

To calculate \( \sum_{k=3}^{5}(-1)^{k} \cdot \frac{2}{k} \), we first expand the sum: \[ \sum_{k=3}^{5}(-1)^{k} \cdot \frac{2}{k} = (-1)^{3} \cdot \frac{2}{3} + (-1)^{4} \cdot \frac{2}{4} + (-1)^{5} \cdot \frac{2}{5} \] Now, calculating each term: 1. For \( k=3 \): \[ (-1)^{3} \cdot \frac{2}{3} = -\frac{2}{3} \] 2. For \( k=4 \): \[ (-1)^{4} \cdot \frac{2}{4} = \frac{2}{4} = \frac{1}{2} \] 3. For \( k=5 \): \[ (-1)^{5} \cdot \frac{2}{5} = -\frac{2}{5} \] Now, we combine these results together: \[ \sum_{k=3}^{5}(-1)^{k} \cdot \frac{2}{k} = -\frac{2}{3} + \frac{1}{2} - \frac{2}{5} \] To add these fractions, we need a common denominator. The least common multiple of 3, 2, and 5 is 30. Now, we convert each fraction: 1. \( -\frac{2}{3} = -\frac{20}{30} \) 2. \( \frac{1}{2} = \frac{15}{30} \) 3. \( -\frac{2}{5} = -\frac{12}{30} \) Combining all terms: \[ -\frac{20}{30} + \frac{15}{30} - \frac{12}{30} = \frac{-20 + 15 - 12}{30} = \frac{-17}{30} \] Thus, the value of the sum is: \[ \boxed{-\frac{17}{30}} \]