The sum `(7)/(2^2xx5^2)+13/(5^2xx8^2)+19/(8^2xx11^2)+…10` terms is S, then the value of 1024(S) is __________.

Explanation

To solve the problem, we need to find the sum \( S \) of the series given by:

\[ S = \frac{7}{2^2 \cdot 5^2} + \frac{13}{5^2 \cdot 8^2} + \frac{19}{8^2 \cdot 11^2} + \ldots \text{ (up to 10 terms)} \]

### Step 1: Identify the general term \( T_n \)

The series can be expressed in terms of a general term \( T_n \). Observing the numerators and denominators, we can see:

- The numerators: 7, 13, 19, ... form an arithmetic progression (AP) with the first term \( a = 7 \) and common difference \( d = 6 \).
- The denominators consist of products of squares of terms that also form an AP.

The \( n \)-th term can be expressed as:

\[ T_n = \frac{7 + 6(n-1)}{(3n-1)^2 \cdot (3n+2)^2} \]

### Step 2: Simplify the general term

The numerator simplifies to:

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

### Step 3: Write the sum \( S \)

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

\[ S = \sum_{n=1}^{10} T_n = \sum_{n=1}^{10} \frac{6n + 1}{(3n-1)^2 \cdot (3n+2)^2} \]

### Step 4: Use the telescoping series

To simplify the sum, we can rewrite \( T_n \) in a form that allows for cancellation:

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

This transformation shows that the series is telescoping.

### Step 5: Calculate the sum

Now we can compute \( S \):

\[ S = \frac{1}{3} \left( \left( \frac{1}{2^2} - \frac{1}{32^2} \right) + \left( \frac{1}{5^2} - \frac{1}{29^2} \right) + \ldots \right) \]

The first term when \( n = 1 \) gives \( \frac{1}{2^2} \) and the last term when \( n = 10 \) gives \( \frac{1}{32^2} \).

### Step 6: Calculate the final value of \( S \)

After performing the calculations, we find:

\[ S = \frac{1}{3} \left( \frac{1}{4} - \frac{1}{1024} \right) \]

Calculating this gives:

\[ S = \frac{1}{3} \left( \frac{256 - 1}{1024} \right) = \frac{255}{3072} \]

### Step 7: Find \( 1024S \)

Finally, we need to find \( 1024S \):

\[ 1024S = 1024 \cdot \frac{255}{3072} = \frac{1024 \cdot 255}{3072} \]

Calculating this gives:

\[ 1024S = 85 \]

Thus, the value of \( 1024(S) \) is:

\[ \boxed{85} \]