`3/(1^2 2^2)+5/(2^2 3^2)+7/(3^2 4^2)+ . . . ` upto 10th term

Answer

To solve the series \( S = \frac{3}{1^2 \cdot 2^2} + \frac{5}{2^2 \cdot 3^2} + \frac{7}{3^2 \cdot 4^2} + \ldots \) up to the 10th term, we first need to identify the general term of the series.

### Step 1: Identify the General Term
The general term \( T_n \) can be expressed as:
\[ T_n = \frac{2n + 1}{n^2 \cdot (n + 1)^2} \]
where \( n \) ranges from 1 to 10.

### Step 2: Rewrite the General Term
We can simplify \( T_n \):
\[ T_n = \frac{2n + 1}{n^2 (n + 1)^2} = \frac{(n + 1)^2 - n^2}{n^2 (n + 1)^2} \]
This can be rewritten as:
\[ T_n = \frac{1}{n^2} - \frac{1}{(n + 1)^2} \]

### Step 3: Sum the Series
Now, we can express the sum \( S \) as:
\[ S = \sum_{n=1}^{10} T_n = \sum_{n=1}^{10} \left( \frac{1}{n^2} - \frac{1}{(n + 1)^2} \right) \]
This is a telescoping series. When we expand it, we get:
\[ S = \left( \frac{1}{1^2} - \frac{1}{2^2} \right) + \left( \frac{1}{2^2} - \frac{1}{3^2} \right) + \left( \frac{1}{3^2} - \frac{1}{4^2} \right) + \ldots + \left( \frac{1}{10^2} - \frac{1}{11^2} \right) \]

### Step 4: Simplify the Sum
In the telescoping series, all intermediate terms cancel out:
\[ S = \frac{1}{1^2} - \frac{1}{11^2} \]
Calculating this gives:
\[ S = 1 - \frac{1}{121} = \frac{121}{121} - \frac{1}{121} = \frac{120}{121} \]

### Final Answer
Thus, the sum of the series up to the 10th term is:
\[ \boxed{\frac{120}{121}} \]