1. **State the problem:** We want to show that $$\sum_{r=3}^n \frac{1}{r(r-2)} = \frac{3}{4} - \frac{1}{2n-2} - \frac{1}{2n}$$ and then find $$\sum_{r=n+1}^{2n} \frac{1}{r(r-2)}$$ 2. **Rewrite the general term using partial fractions:** We start by expressing $$\frac{1}{r(r-2)}$$ as $$\frac{A}{r-2} + \frac{B}{r}$$ Multiply both sides by $r(r-2)$: $$1 = A r + B (r-2) = (A + B) r - 2B$$ Equate coefficients: - Coefficient of $r$: $A + B = 0$ - Constant term: $-2B = 1 \Rightarrow B = -\frac{1}{2}$ Then $A = -B = \frac{1}{2}$. So, $$\frac{1}{r(r-2)} = \frac{1/2}{r-2} - \frac{1/2}{r} = \frac{1}{2} \left( \frac{1}{r-2} - \frac{1}{r} \right)$$ 3. **Write the sum using this decomposition:** $$\sum_{r=3}^n \frac{1}{r(r-2)} = \frac{1}{2} \sum_{r=3}^n \left( \frac{1}{r-2} - \frac{1}{r} \right) = \frac{1}{2} \left( \sum_{r=3}^n \frac{1}{r-2} - \sum_{r=3}^n \frac{1}{r} \right)$$ 4. **Change indices to simplify sums:** For the first sum, $$\sum_{r=3}^n \frac{1}{r-2} = \sum_{k=1}^{n-2} \frac{1}{k}$$ For the second sum, $$\sum_{r=3}^n \frac{1}{r} = \sum_{k=3}^n \frac{1}{k}$$ So, $$\sum_{r=3}^n \frac{1}{r(r-2)} = \frac{1}{2} \left( \sum_{k=1}^{n-2} \frac{1}{k} - \sum_{k=3}^n \frac{1}{k} \right)$$ 5. **Combine sums:** Write explicitly: $$\sum_{k=1}^{n-2} \frac{1}{k} = 1 + \frac{1}{2} + \sum_{k=3}^{n-2} \frac{1}{k}$$ $$\sum_{k=3}^n \frac{1}{k} = \sum_{k=3}^{n-2} \frac{1}{k} + \frac{1}{n-1} + \frac{1}{n}$$ Subtracting, $$\sum_{k=1}^{n-2} \frac{1}{k} - \sum_{k=3}^n \frac{1}{k} = \left(1 + \frac{1}{2} + \sum_{k=3}^{n-2} \frac{1}{k} \right) - \left( \sum_{k=3}^{n-2} \frac{1}{k} + \frac{1}{n-1} + \frac{1}{n} \right) = 1 + \frac{1}{2} - \frac{1}{n-1} - \frac{1}{n}$$ 6. **Substitute back:** $$\sum_{r=3}^n \frac{1}{r(r-2)} = \frac{1}{2} \left( 1 + \frac{1}{2} - \frac{1}{n-1} - \frac{1}{n} \right) = \frac{1}{2} \times \frac{3}{2} - \frac{1}{2} \left( \frac{1}{n-1} + \frac{1}{n} \right) = \frac{3}{4} - \frac{1}{2(n-1)} - \frac{1}{2n}$$ Rewrite denominators: $$\frac{1}{2(n-1)} = \frac{1}{2n - 2}$$ So, $$\sum_{r=3}^n \frac{1}{r(r-2)} = \frac{3}{4} - \frac{1}{2n - 2} - \frac{1}{2n}$$ This proves the first part. 7. **Find** $$\sum_{r=n+1}^{2n} \frac{1}{r(r-2)}$$ Use the telescoping property: $$\sum_{r=3}^{2n} \frac{1}{r(r-2)} = \frac{3}{4} - \frac{1}{4n - 2} - \frac{1}{4n}$$ and $$\sum_{r=3}^n \frac{1}{r(r-2)} = \frac{3}{4} - \frac{1}{2n - 2} - \frac{1}{2n}$$ Therefore, $$\sum_{r=n+1}^{2n} \frac{1}{r(r-2)} = \sum_{r=3}^{2n} \frac{1}{r(r-2)} - \sum_{r=3}^n \frac{1}{r(r-2)} = \left( \frac{3}{4} - \frac{1}{4n - 2} - \frac{1}{4n} \right) - \left( \frac{3}{4} - \frac{1}{2n - 2} - \frac{1}{2n} \right)$$ Simplify: $$= - \frac{1}{4n - 2} - \frac{1}{4n} + \frac{1}{2n - 2} + \frac{1}{2n}$$ Rewrite denominators: - $4n - 2 = 2(2n - 1)$ - $2n - 2 = 2(n - 1)$ So, $$= \frac{1}{2(n-1)} + \frac{1}{2n} - \frac{1}{2(2n - 1)} - \frac{1}{4n}$$ Combine terms carefully to get the simplified expression. **Final answers:** $$\sum_{r=3}^n \frac{1}{r(r-2)} = \frac{3}{4} - \frac{1}{2n - 2} - \frac{1}{2n}$$ $$\sum_{r=n+1}^{2n} \frac{1}{r(r-2)} = \frac{1}{2(n-1)} + \frac{1}{2n} - \frac{1}{2(2n - 1)} - \frac{1}{4n}$$