Question: Find a formula for
$\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + ... + \frac{1}{n(n+1)}$
by examining the values of this expression for small
values of $n$.
Prove the formula you conjectured in part (a).
1. **State the problem:**
We want to find a formula for the sum
$$S_n = \frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \cdots + \frac{1}{n(n+1)}$$
for positive integers $n$.
2. **Examine small values:**
- For $n=1$: $S_1 = \frac{1}{1\cdot 2} = \frac{1}{2}$
- For $n=2$: $S_2 = \frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} = \frac{1}{2} + \frac{1}{6} = \frac{2}{3}$
- For $n=3$: $S_3 = \frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} = \frac{3}{4}$
- For $n=4$: $S_4 = \frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} + \frac{1}{4\cdot 5} = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} = \frac{4}{5}$
3. **Conjecture the formula:**
From the pattern, it appears that
$$S_n = \frac{n}{n+1}$$
4. **Prove the formula by induction:**
- **Base case:** For $n=1$,
$$S_1 = \frac{1}{1\cdot 2} = \frac{1}{2} = \frac{1}{1+1}$$
which matches the formula.
- **Inductive step:** Assume the formula holds for $n=k$, i.e.,
$$S_k = \sum_{i=1}^k \frac{1}{i(i+1)} = \frac{k}{k+1}$$
We want to prove it for $n=k+1$:
$$S_{k+1} = S_k + \frac{1}{(k+1)(k+2)}$$
Using the inductive hypothesis:
$$S_{k+1} = \frac{k}{k+1} + \frac{1}{(k+1)(k+2)}$$
5. **Simplify the right side:**
$$\frac{k}{k+1} + \frac{1}{(k+1)(k+2)} = \frac{k(k+2)}{(k+1)(k+2)} + \frac{1}{(k+1)(k+2)} = \frac{k(k+2) + 1}{(k+1)(k+2)}$$
6. **Calculate numerator:**
$$k(k+2) + 1 = k^2 + 2k + 1 = (k+1)^2$$
7. **Substitute back:**
$$S_{k+1} = \frac{(k+1)^2}{(k+1)(k+2)} = \frac{k+1}{k+2}$$
8. **Conclusion:**
By induction, the formula
$$\sum_{i=1}^n \frac{1}{i(i+1)} = \frac{n}{n+1}$$
holds for all positive integers $n$.
**Final answer:**
$$\boxed{\sum_{i=1}^n \frac{1}{i(i+1)} = \frac{n}{n+1}}$$