The problem is asking to calculate the sum of the series:

$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \cdots + \frac{1}{149 \cdot 150} + \frac{1}{150 \cdot 151}.$

This series can be simplified using partial fraction decomposition.

Step-by-Step Solution:

1. General Term Simplification: Each term in the series can be expressed as: $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}.$

2. Substitute in the Series: The series becomes: $\left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \cdots + \left( \frac{1}{150} - \frac{1}{151} \right).$

3. Telescoping Effect: Most terms cancel out, leaving: $\frac{1}{1} - \frac{1}{151}.$

4. Final Calculation: $1 - \frac{1}{151} = \frac{151}{151} - \frac{1}{151} = \frac{150}{151}.$

Final Answer:

The sum of the series is: $\boxed{\frac{150}{151}}$

Would you like further clarification or a detailed breakdown of any part of the solution?

Related Questions:

1. How does partial fraction decomposition work in general?
2. Can this telescoping technique be applied to other series?
3. What happens if the upper limit of the series is extended to infinity?
4. How would this series change if the denominator was $n(n+k)$ instead of $n(n+1)$?
5. What are the practical applications of such telescoping series in real-world problems?

Tip:

When solving series problems, always check for patterns (like telescoping) to simplify computations efficiently.