Sum to n terms the series
`(3)/(1 ^(2) . 2 ^(2)) + (5)/( 2 ^(2) . 3 ^(2)) + (7)/( 3 ^(2) . 4 ^(2)) +.............`

Answer

To find the sum to n terms of the series

\[ S_n = \frac{3}{1^2 \cdot 2^2} + \frac{5}{2^2 \cdot 3^2} + \frac{7}{3^2 \cdot 4^2} + \ldots \]

we can analyze the general term of the series.

### Step 1: Identify the general term

The general term of the series can be expressed as:

\[ T_n = \frac{2n + 1}{n^2 \cdot (n + 1)^2} \]

This is because:
- The numerator \(2n + 1\) corresponds to the sequence \(3, 5, 7, \ldots\), which can be represented as \(2n + 1\) for \(n = 1, 2, 3, \ldots\).
- The denominator consists of \(n^2\) and \((n + 1)^2\).

### Step 2: Rewrite the general term

We can rewrite the general term \(T_n\) as:

\[ T_n = \frac{(n + 1)^2 - n^2}{n^2 \cdot (n + 1)^2} \]

This simplifies to:

\[ T_n = \frac{(n + 1)^2 - n^2}{n^2 \cdot (n + 1)^2} = \frac{1}{n^2} - \frac{1}{(n + 1)^2} \]

### Step 3: Sum the series

Now, we can sum the series \(S_n\):

\[ S_n = \sum_{k=1}^{n} \left(\frac{1}{k^2} - \frac{1}{(k + 1)^2}\right) \]

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

\[ S_n = \left(1 - \frac{1}{2^2}\right) + \left(\frac{1}{2^2} - \frac{1}{3^2}\right) + \left(\frac{1}{3^2} - \frac{1}{4^2}\right) + \ldots + \left(\frac{1}{n^2} - \frac{1}{(n + 1)^2}\right) \]

### Step 4: Simplify the sum

In a telescoping series, all intermediate terms cancel out, leaving:

\[ S_n = 1 - \frac{1}{(n + 1)^2} \]

### Step 5: Final expression for \(S_n\)

Thus, the sum to n terms of the series is:

\[ S_n = 1 - \frac{1}{(n + 1)^2} \]