If \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\dots+\frac{1}{n(n+1)}=\frac{99}{100}, then what is the value of n?

Explanation

Using partial fractions, \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}. The series forms a telescoping sum that simplifies to 1 - \frac{1}{n+1} = \frac{n}{n+1}. Setting \frac{n}{n+1} = \frac{99}{100} gives n = 99.

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