`1/(3^2-1)+1/(5^2-1)+1/(7^2-1)+ . . . + 1/(201^2-1)` is equal to

Answer

To solve the series

\[ S = \frac{1}{3^2 - 1} + \frac{1}{5^2 - 1} + \frac{1}{7^2 - 1} + \ldots + \frac{1}{201^2 - 1}, \]

we first notice that the terms in the series involve the squares of odd numbers. The odd numbers can be expressed in the form \(2n + 1\), where \(n\) is a non-negative integer. The first odd number is 3, which corresponds to \(n = 1\), and the last odd number is 201, which corresponds to \(n = 100\).

### Step 1: Rewrite the series using summation notation

The series can be rewritten as:

\[ S = \sum_{n=1}^{100} \frac{1}{(2n + 1)^2 - 1}. \]

### Step 2: Simplify the denominator

Next, we simplify the term in the denominator:

\[ (2n + 1)^2 - 1 = 4n^2 + 4n + 1 - 1 = 4n^2 + 4n. \]

Thus, we can rewrite the series as:

\[ S = \sum_{n=1}^{100} \frac{1}{4n^2 + 4n} = \sum_{n=1}^{100} \frac{1}{4n(n + 1)}. \]

### Step 3: Factor out the constant

We can factor out the constant \( \frac{1}{4} \):

\[ S = \frac{1}{4} \sum_{n=1}^{100} \frac{1}{n(n + 1)}. \]

### Step 4: Use partial fractions

Next, we can decompose \(\frac{1}{n(n + 1)}\) using partial fractions:

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

### Step 5: Rewrite the sum

Now we can rewrite the sum:

\[ S = \frac{1}{4} \sum_{n=1}^{100} \left( \frac{1}{n} - \frac{1}{n + 1} \right). \]

### Step 6: Recognize the telescoping series

This is a telescoping series. When we expand it, we get:

\[ S = \frac{1}{4} \left( \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \ldots + \left( \frac{1}{100} - \frac{1}{101} \right) \right). \]

### Step 7: Simplify the series

Most terms will cancel out, leaving us with:

\[ S = \frac{1}{4} \left( 1 - \frac{1}{101} \right) = \frac{1}{4} \left( \frac{101 - 1}{101} \right) = \frac{1}{4} \cdot \frac{100}{101} = \frac{25}{101}. \]

### Final Result

Thus, the sum of the series is:

\[ S = \frac{25}{101}. \]