1. **Problem 10a:** Find a formula for the sum $$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{n(n+1)}$$ 2. **Step 1: Examine small values of $n$** - For $n=1$: $\frac{1}{1\cdot 2} = \frac{1}{2}$ - For $n=2$: $\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} = \frac{1}{2} + \frac{1}{6} = \frac{2}{3}$ - For $n=3$: $\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} = \frac{3}{4}$ 3. **Step 2: Conjecture a formula** Notice the pattern: $\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \ldots$ which suggests $$\sum_{k=1}^n \frac{1}{k(k+1)} = \frac{n}{n+1}$$ 4. **Step 3: Prove the formula (Problem 10b)** Use partial fraction decomposition: $$\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$$ So the sum becomes a telescoping series: $$\sum_{k=1}^n \left( \frac{1}{k} - \frac{1}{k+1} \right) = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right)$$ All intermediate terms cancel, leaving: $$1 - \frac{1}{n+1} = \frac{n}{n+1}$$ which confirms the conjectured formula. --- 5. **Problem 11a:** Find a formula for the sum $$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots + \frac{1}{2^n}$$ 6. **Step 1: Examine small values of $n$** - For $n=1$: $\frac{1}{2} = 0.5$ - For $n=2$: $\frac{1}{2} + \frac{1}{4} = 0.75$ - For $n=3$: $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} = 0.875$ 7. **Step 2: Conjecture a formula** This is a geometric series with first term $a=\frac{1}{2}$ and ratio $r=\frac{1}{2}$. The sum of the first $n$ terms is: $$S_n = a \frac{1-r^n}{1-r} = \frac{1}{2} \frac{1 - \left(\frac{1}{2}\right)^n}{1 - \frac{1}{2}} = \frac{1}{2} \frac{1 - \frac{1}{2^n}}{\frac{1}{2}} = 1 - \frac{1}{2^n}$$ 8. **Step 3: Prove the formula (Problem 11b)** By the formula for the sum of a geometric series: $$S_n = a \frac{1-r^n}{1-r}$$ Substitute $a=\frac{1}{2}$ and $r=\frac{1}{2}$: $$S_n = \frac{1}{2} \frac{1 - \left(\frac{1}{2}\right)^n}{1 - \frac{1}{2}} = 1 - \frac{1}{2^n}$$ This matches the values found and proves the formula. **Final answers:** - For problem 10: $$\sum_{k=1}^n \frac{1}{k(k+1)} = \frac{n}{n+1}$$ - For problem 11: $$\sum_{k=1}^n \frac{1}{2^k} = 1 - \frac{1}{2^n}$$