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Solution: We begin by simplifying the general term using partial fractions:
\frac{1}{n(n+2)} = \frac{A}{n} + \frac{B}{n+2}.
Multiplying both sides by $ n(n+2) $:
= A(n+2) + Bn.
= An + 2A + Bn = (A + B)n + 2A.
A + B = 0, \quad 2A = 1 \Rightarrow A = \frac{1}{2}, \quad B = -\frac{1}{2}.
\frac{1}{n(n+2)} = \frac{1}{2} \left( \frac{1}{n} - \frac{1}{n+2} \right).
\sum_{n=1}^{50} \frac{1}{n(n+2)} = \frac{1}{2} \sum_{n=1}^{50} \left( \frac{1}{n} - \frac{1}{n+2} \right).
This is a telescoping sum. Write out the first few terms:
\left( \frac{1}{1} - \frac{1}{3} \right) + \left( \frac{1}{2} - \frac{1}{4} \right) + \left( \frac{1}{3} - \frac{1}{5} \right) + \cdots + \left( \frac{1}{50} - \frac{1}{52} \right).