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To solve this problem, let's break down both parts (a) and (b) of the question.

### Part (a): Find a Formula for the Series

The series we are given is:

\[
\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots + \frac{1}{n(n+1)}
\]

#### Step 1: Compute the First Few Terms

Let's evaluate the partial sums for small values of \( n \) to observe a possible pattern.

1. **For \( n = 1 \):**
   \[
   \frac{1}{1 \cdot 2} = \frac{1}{2}
   \]

2. **For \( n = 2 \):**
   \[
   \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} = \frac{1}{2} + \frac{1}{6} = \frac{2 + 1}{3} = \frac{2}{3}
   \]

3. **For \( n = 3 \):**
   \[
   \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} = \frac{3 + 1}{4} = \frac{3}{4}
   \]

4. **For \( n = 4 \):**
   \[
   \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \frac{1}{4 \cdot 5} = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} = \frac{4}{5}
   \]

From this, it appears that the partial sum for the series up to \( n \) terms is:

\[
S_n = 1 - \frac{1}{n+1}
\]

Thus, the formula we conjecture for the sum of the series up to \( n \) terms is:

\[
\sum_{k=1}^n \frac{1}{k(k+1)} = 1 - \frac{1}{n+1}
\]

---

### Part (b): Prove the Formula Conjectured in Part (a)

To prove this formula, we can use **partial fraction decomposition** to rewrite the general term \( \frac{1}{k(k+1)} \).

#### Step 1: Partial Fraction Decomposition

Rewrite \( \frac{1}{k(k+1)} \) as a difference of two fractions:

\[
\frac{1}{k(k+1)} = \frac{A}{k} + \frac{B}{k+1}
\]

Multiplying through by \( k(k+1) \) gives:

\[
1 = A(k+1) + Bk
\]

Setting up equations by expanding and matching terms:

1. For the constant term: \( A = 1 \)
2. For the \( k \) term: \( A + B = 0 \), so \( B = -1 \)

Thus:

\[
\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}
\]

#### Step 2: Rewrite the Series Using Partial Fractions

Substituting the partial fraction decomposition into the series:

\[
\sum_{k=1}^n \frac{1}{k(k+1)} = \sum_{k=1}^n \left( \frac{1}{k} - \frac{1}{k+1} \right)
\]

#### Step 3: Observe the Telescoping Effect

The series now becomes:

\[
\left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \cdots + \left( \frac{1}{n} - \frac{1}{n+1} \right)
\]

Most terms cancel out, leaving only:

\[
1 - \frac{1}{n+1}
\]

#### Conclusion

Thus, we have shown that:

\[
\sum_{k=1}^n \frac{1}{k(k+1)} = 1 - \frac{1}{n+1}
\]

This confirms the conjectured formula.

---

Would you like a further breakdown of any step? 

Here are some related questions you might explore:
1. How does partial fraction decomposition help simplify series?
2. Can this technique be applied to other series involving products in denominators?
3. How does telescoping work in summing series?
4. What are other methods to find closed forms for series?
5. How can induction be used to prove the validity of series formulas?

**Tip:** For series involving fractions, check if partial fraction decomposition can simplify the terms into a telescoping sum.

Find a formula for \( \sum_{k=1}^n \frac{1}{k(k+1)} \) and prove it.

Learn how to find and prove the formula for the sum of the series \( \sum_{k=1}^n \frac{1}{k(k+1)} \) using partial fraction decomposition and telescoping series techniques. Suitable for advanced high school students in Grades 11-12.

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