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The problem defines a sequence \( a_n \) with the formula:

\[
a_n = \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{n(n+1)}
\]

This type of series is telescoping, where each term can be simplified to reveal a pattern that makes summing easier.

To evaluate \( a_n \), let’s break down each term in the series:

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

Using partial fraction decomposition, each term in the series becomes:

\[
\frac{1}{1 \cdot 2} = \frac{1}{1} - \frac{1}{2}, \quad \frac{1}{2 \cdot 3} = \frac{1}{2} - \frac{1}{3}, \quad \text{and so on,} \quad \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}
\]

So, the sequence \( a_n \) becomes:

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

In this telescoping series, most terms cancel out, leaving only:

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

Therefore,

\[
a_n = \frac{n}{n+1}
\]

Would you like further details on any step or have any questions?

Here are five related questions that build on this solution:

1. How would you find the limit of \( a_n \) as \( n \to \infty \)?
2. Can you express \( a_n \) as a continuous function of \( n \)?
3. How would the series change if the general term was \( \frac{1}{n(n+2)} \)?
4. What happens if the sum goes to infinity? Would it converge or diverge?
5. How could you verify the solution by induction?

**Tip:** Telescoping series often simplify complex sums by breaking terms down into parts that cancel sequentially.

Evaluate the series a_n = 1/(1*2) + 1/(2*3) + ... + 1/(n(n+1))

Discover how to solve the telescoping series a_n = 1/(1*2) + 1/(2*3) + ... + 1/(n(n+1)) by simplifying each term with partial fractions. This solution breaks down the process step-by-step, suitable for high school students in Grades 10-12.

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