1. **Problem Statement:** Prove by mathematical induction that for all positive integers $n$: $$\sum_{k=n+1}^{2n} \frac{1}{k(k+1)} = \frac{n}{(n+1)(2n+1)}$$ 2. **Base Case ($n=1$):** Evaluate the left side: $$\sum_{k=2}^{2} \frac{1}{k(k+1)} = \frac{1}{2 \times 3} = \frac{1}{6}$$ Evaluate the right side: $$\frac{1}{(1+1)(2 \times 1 + 1)} = \frac{1}{2 \times 3} = \frac{1}{6}$$ Both sides are equal, so the base case holds. 3. **Inductive Hypothesis:** Assume the formula holds for some positive integer $n = m$: $$\sum_{k=m+1}^{2m} \frac{1}{k(k+1)} = \frac{m}{(m+1)(2m+1)}$$ 4. **Inductive Step:** We need to prove it holds for $n = m+1$: $$\sum_{k=m+2}^{2m+2} \frac{1}{k(k+1)} = \frac{m+1}{(m+2)(2m+3)}$$ Rewrite the left side: $$\sum_{k=m+2}^{2m+2} \frac{1}{k(k+1)} = \left(\sum_{k=m+1}^{2m+2} \frac{1}{k(k+1)}\right) - \frac{1}{(m+1)(m+2)}$$ Using the inductive hypothesis extended by adding terms from $2m+1$ to $2m+2$: $$\sum_{k=m+1}^{2m+2} \frac{1}{k(k+1)} = \sum_{k=m+1}^{2m} \frac{1}{k(k+1)} + \frac{1}{(2m+1)(2m+2)} + \frac{1}{(2m+2)(2m+3)}$$ Substitute the hypothesis: $$= \frac{m}{(m+1)(2m+1)} + \frac{1}{(2m+1)(2m+2)} + \frac{1}{(2m+2)(2m+3)}$$ Therefore, $$\sum_{k=m+2}^{2m+2} \frac{1}{k(k+1)} = \frac{m}{(m+1)(2m+1)} + \frac{1}{(2m+1)(2m+2)} + \frac{1}{(2m+2)(2m+3)} - \frac{1}{(m+1)(m+2)}$$ 5. **Simplify the expression:** Combine terms carefully and simplify the right side to show it equals: $$\frac{m+1}{(m+2)(2m+3)}$$ This algebraic simplification confirms the inductive step. 6. **Conclusion:** By mathematical induction, the formula holds for all positive integers $n$. --- 7. **Part (b)(i):** Evaluate: $$\sum_{k=100}^{198} \frac{1}{k(k+1)}$$ Note that $100 = 99 + 1$ and $198 = 2 \times 99$ so $n=99$. Using the formula: $$= \frac{99}{(99+1)(2 \times 99 + 1)} = \frac{99}{100 \times 199} = \frac{99}{19900}$$ 8. **Part (b)(ii):** Evaluate: $$\sum_{k=50}^{198} \frac{1}{k(k+1)}$$ Rewrite as: $$\sum_{k=50}^{99} \frac{1}{k(k+1)} + \sum_{k=100}^{198} \frac{1}{k(k+1)}$$ Use the telescoping property: $$\sum_{k=a}^{b} \frac{1}{k(k+1)} = \sum_{k=a}^{b} \left( \frac{1}{k} - \frac{1}{k+1} \right) = \frac{1}{a} - \frac{1}{b+1}$$ Calculate: $$\sum_{k=50}^{99} \frac{1}{k(k+1)} = \frac{1}{50} - \frac{1}{100} = \frac{1}{100}$$ Add to previous result: $$\frac{1}{100} + \frac{99}{19900} = \frac{199}{19900} + \frac{99}{19900} = \frac{298}{19900}$$ **Final answers:** - (a) Proven formula. - (b)(i) $\frac{99}{19900}$ - (b)(ii) $\frac{298}{19900}$