1. **State the problem:** Calculate the sum $$\sum_{j=0}^4 \frac{j}{j+2}$$. 2. **Formula and explanation:** This is a finite sum where each term is a fraction with numerator $j$ and denominator $j+2$. We will evaluate each term individually and then add them up. 3. **Calculate each term:** - For $j=0$: $$\frac{0}{0+2} = \frac{0}{2} = 0$$ - For $j=1$: $$\frac{1}{1+2} = \frac{1}{3}$$ - For $j=2$: $$\frac{2}{2+2} = \frac{2}{4} = \frac{1}{2}$$ - For $j=3$: $$\frac{3}{3+2} = \frac{3}{5}$$ - For $j=4$: $$\frac{4}{4+2} = \frac{4}{6} = \frac{2}{3}$$ 4. **Sum all terms:** $$0 + \frac{1}{3} + \frac{1}{2} + \frac{3}{5} + \frac{2}{3}$$ 5. **Find common denominator and add:** The denominators are 3, 2, 5, and 3. The least common denominator (LCD) is 30. Convert each fraction: - $\frac{1}{3} = \frac{10}{30}$ - $\frac{1}{2} = \frac{15}{30}$ - $\frac{3}{5} = \frac{18}{30}$ - $\frac{2}{3} = \frac{20}{30}$ Sum: $$0 + \frac{10}{30} + \frac{15}{30} + \frac{18}{30} + \frac{20}{30} = \frac{10 + 15 + 18 + 20}{30} = \frac{63}{30}$$ 6. **Simplify the fraction:** $$\frac{63}{30} = \frac{\cancel{3}21}{\cancel{3}10} = \frac{21}{10} = 2.1$$ **Final answer:** $$\sum_{j=0}^4 \frac{j}{j+2} = \frac{21}{10} = 2.1$$