$$\frac{1}{3^2 - 1} + \frac{1}{5^2 - 1} + \frac{1}{7^2 - 1} + \ldots + \frac{1}{(201)^2 - 1}$$ is equal to:

We use partial fractions. Each term has the form $$\frac{1}{n^2 - 1} = \frac{1}{(n-1)(n+1)} = \frac{1}{2}\left(\frac{1}{n-1} - \frac{1}{n+1}\right)$$.

The series is $$\sum_{k=1}^{100} \frac{1}{(2k+1)^2 - 1}$$ where $$n = 2k+1$$ runs through 3, 5, 7, ..., 201. So the sum equals $$\frac{1}{2}\sum_{k=1}^{100}\left(\frac{1}{2k} - \frac{1}{2k+2}\right) = \frac{1}{4}\sum_{k=1}^{100}\left(\frac{1}{k} - \frac{1}{k+1}\right)$$.

This is a telescoping series: $$\frac{1}{4}\left(\frac{1}{1} - \frac{1}{101}\right) = \frac{1}{4} \cdot \frac{100}{101} = \frac{100}{404} = \frac{25}{101}$$.

The answer is Option B: $$\frac{25}{101}$$.