The sum of the series `(5)/(1^(2).4^(2)) + (11)/(4^(2) . 7^(2)) + (17)/(7^(2) . 10^(2))+..` ad inf. is `(1)/(3)`.

Answer

To solve the series

\[ S = \frac{5}{1^2 \cdot 4^2} + \frac{11}{4^2 \cdot 7^2} + \frac{17}{7^2 \cdot 10^2} + \ldots \]

we will analyze the general term of the series and find its sum.

### Step 1: Identify the general term

The numerators of the series are \(5, 11, 17, \ldots\). This sequence can be expressed as:

\[ a_n = 5 + (n-1) \cdot 6 = 6n - 1 \]

The denominators consist of products of squares of numbers in an arithmetic progression. The first term is \(1\), the second term is \(4\), the third term is \(7\), and so on. This sequence can be expressed as:

\[ b_n = 3n - 2 \quad \text{(for the first term)} \]
\[ c_n = 3n + 1 \quad \text{(for the second term)} \]

Thus, the general term \(P_n\) of the series can be written as:

\[ P_n = \frac{6n - 1}{(3n - 2)^2 (3n + 1)^2} \]

### Step 2: Simplify the general term

We can rewrite \(P_n\) as follows:

\[ P_n = \frac{6n - 1}{(3n - 2)^2 (3n + 1)^2} = \frac{1}{3} \left( \frac{1}{(3n - 2)^2} - \frac{1}{(3n + 1)^2} \right) \]

This is achieved by recognizing that the numerator \(6n - 1\) can be expressed as the difference of two squares.

### Step 3: Sum the series

Now we can express the sum \(S\) as:

\[ S = \sum_{n=1}^{\infty} P_n = \frac{1}{3} \sum_{n=1}^{\infty} \left( \frac{1}{(3n - 2)^2} - \frac{1}{(3n + 1)^2} \right) \]

This series is telescoping. When we write out the first few terms, we can see that most terms will cancel:

\[ S = \frac{1}{3} \left( \left( \frac{1}{1^2} - \frac{1}{4^2} \right) + \left( \frac{1}{4^2} - \frac{1}{7^2} \right) + \left( \frac{1}{7^2} - \frac{1}{10^2} \right) + \ldots \right) \]

### Step 4: Evaluate the limit

As \(n\) approaches infinity, the last term \(\frac{1}{(3n + 1)^2}\) approaches zero. Thus, we are left with:

\[ S = \frac{1}{3} \left( 1 - 0 \right) = \frac{1}{3} \]

### Conclusion

Therefore, the sum of the series is

\[ S = \frac{1}{3} \]