Evaluate: \( \quad \sum_{k=3}^{5}\left[(-1)^{k} \times \frac{2}{k}\right] \)

To evaluate the sum \( \sum_{k=3}^{5}\left[(-1)^{k} \times \frac{2}{k}\right] \), we will compute the individual terms for \( k = 3, 4, 5 \) and then sum them up. We start with \( k = 3 \): \[ (-1)^3 \times \frac{2}{3} = -1 \times \frac{2}{3} = -\frac{2}{3} \] Next, for \( k = 4 \): \[ (-1)^4 \times \frac{2}{4} = 1 \times \frac{2}{4} = \frac{1}{2} \] Finally, for \( k = 5 \): \[ (-1)^5 \times \frac{2}{5} = -1 \times \frac{2}{5} = -\frac{2}{5} \] Now, we combine all these results into the total sum: \[ \sum_{k=3}^{5} \left[ (-1)^{k} \times \frac{2}{k} \right] = -\frac{2}{3} + \frac{1}{2} - \frac{2}{5} \] To compute this sum, we need a common denominator. The least common multiple of the denominators \( 3, 2, \) and \( 5 \) is \( 30 \). Now, we express each term with a denominator of \( 30 \): - For \( -\frac{2}{3} \): \[ -\frac{2}{3} = -\frac{2 \times 10}{3 \times 10} = -\frac{20}{30} \] - For \( \frac{1}{2} \): \[ \frac{1}{2} = \frac{1 \times 15}{2 \times 15} = \frac{15}{30} \] - For \( -\frac{2}{5} \): \[ -\frac{2}{5} = -\frac{2 \times 6}{5 \times 6} = -\frac{12}{30} \] Now we can add these fractions: \[ -\frac{20}{30} + \frac{15}{30} - \frac{12}{30} = \frac{-20 + 15 - 12}{30} = \frac{-17}{30} \] Thus, the final result of the sum is: \[ \boxed{-\frac{17}{30}} \]