the sum`3/1^2+5/(1^2+2^2)+7/(1^2+2^2+3^2)+.......` upto 11 terms

Options

A. `(11)/(4)`
B. `(60)/(11)`
C. `(7)/(2)`
D. `(11)/(2)`

Answer

To solve the series \( S = \frac{3}{1^2} + \frac{5}{1^2 + 2^2} + \frac{7}{1^2 + 2^2 + 3^2} + \ldots \) up to 11 terms, we can follow these steps:

### Step 1: Identify the General Term
The general term \( T_n \) of the series can be expressed as:
\[ T_n = \frac{2n + 1}{1^2 + 2^2 + 3^2 + \ldots + n^2} \]
where \( n \) is the term number.

### Step 2: Sum of Squares Formula
The sum of the squares of the first \( n \) natural numbers is given by:
\[ 1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{n(n + 1)(2n + 1)}{6} \]
Thus, we can rewrite the general term:
\[ T_n = \frac{2n + 1}{\frac{n(n + 1)(2n + 1)}{6}} = \frac{6(2n + 1)}{n(n + 1)(2n + 1)} \]
This simplifies to:
\[ T_n = \frac{6}{n(n + 1)} \]

### Step 3: Partial Fraction Decomposition
Next, we can decompose \( T_n \):
\[ T_n = \frac{6}{n(n + 1)} = 6 \left( \frac{1}{n} - \frac{1}{n + 1} \right) \]

### Step 4: Write the Sum of the Series
Now, we can write the sum \( S \) of the first 11 terms:
\[ S = \sum_{n=1}^{11} T_n = \sum_{n=1}^{11} 6 \left( \frac{1}{n} - \frac{1}{n + 1} \right) \]
This is a telescoping series.

### Step 5: Evaluate the Sum
When we expand the sum, we have:
\[ S = 6 \left( \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \ldots + \left( \frac{1}{11} - \frac{1}{12} \right) \right) \]
Most terms cancel out, leaving us with:
\[ S = 6 \left( 1 - \frac{1}{12} \right) = 6 \left( \frac{12 - 1}{12} \right) = 6 \left( \frac{11}{12} \right) = \frac{66}{12} = \frac{11}{2} \]

### Final Answer
Thus, the sum of the series up to 11 terms is:
\[ \boxed{\frac{11}{2}} \]