`5/(1^2*4^2)+11/(4^2*7^2)+17/(7^2*1 0^2)+`

Options

A. `(1)/(2)`
B. `(1)/(3)`
C. 2
D. 3

Answer

To solve the given 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 pattern in the numerators and denominators.

### Step 1: Identify the pattern in the numerators
The numerators are:
- 5 can be expressed as \(1 + 4\)
- 11 can be expressed as \(4 + 7\)
- 17 can be expressed as \(7 + 10\)

This suggests that the \(n\)-th term's numerator can be expressed as the sum of two consecutive terms in the sequence: \(a_n + a_{n+1}\), where \(a_n\) is the sequence of numbers \(1, 4, 7, 10, \ldots\) which can be defined as \(a_n = 1 + 3(n-1)\).

### Step 2: Identify the pattern in the denominators
The denominators are products of squares:
- The first term has \(1^2 \cdot 4^2\)
- The second term has \(4^2 \cdot 7^2\)
- The third term has \(7^2 \cdot 10^2\)

This suggests that the \(n\)-th term's denominator can be expressed as \(a_n^2 \cdot a_{n+1}^2\).

### Step 3: Rewrite the series
We can rewrite the series as follows:

\[ S = \sum_{n=1}^{\infty} \frac{a_n + a_{n+1}}{a_n^2 \cdot a_{n+1}^2} \]

### Step 4: Simplify the terms
We can separate the terms in the summation:

\[ S = \sum_{n=1}^{\infty} \left( \frac{1}{a_n \cdot a_{n+1}^2} + \frac{1}{a_{n+1} \cdot a_n^2} \right) \]

### Step 5: Use the difference of squares
We can express the terms using the difference of squares:

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

### Step 6: Apply telescoping series
Notice that the series has a telescoping nature. When we expand the series, many terms will cancel out.

### Step 7: Evaluate the limit
As \(n\) approaches infinity, the remaining terms will converge to a finite value. The limit of the series can be evaluated as:

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

### Final Answer
Thus, the sum of the series is:

\[ \boxed{\frac{1}{3}} \]