The sum up to 60 terms of `3/(1^2) + 5/(1^2 + 2^2) + 7/(1^2 + 2^2 + 3^2) + …….` is equal to

Options

A. `240/61`
B. `180/17`
C. `360/61`
D. `100/17`

Answer

To find the sum of the series up to 60 terms given by:

\[ S = \frac{3}{1^2} + \frac{5}{1^2 + 2^2} + \frac{7}{1^2 + 2^2 + 3^2} + \ldots \]

we will follow these steps:

### Step 1: Identify the general term
The numerators of the terms are in an arithmetic progression (AP): 3, 5, 7, ..., which can be expressed as:
\[ a_n = 2n + 1 \quad \text{(for } n \text{ starting from 1)} \]

### Step 2: Determine the denominator
The denominators are the sums of squares:
\[ 1^2 + 2^2 + \ldots + n^2 \]
The formula for the sum of the first \( n \) squares is:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]

### Step 3: Write the \( n \)-th term
Thus, the \( n \)-th term of the series can be written as:
\[ T_n = \frac{2n + 1}{\frac{n(n + 1)(2n + 1)}{6}} = \frac{6(2n + 1)}{n(n + 1)(2n + 1)} \]
This simplifies to:
\[ T_n = \frac{6}{n(n + 1)} \]

### Step 4: Simplify the term
We can further simplify \( T_n \):
\[ T_n = 6 \left( \frac{1}{n} - \frac{1}{n + 1} \right) \]

### Step 5: Sum the series
Now, we need to find the sum of the first 60 terms:
\[ S_{60} = \sum_{n=1}^{60} T_n = \sum_{n=1}^{60} 6 \left( \frac{1}{n} - \frac{1}{n + 1} \right) \]
This is a telescoping series. When we expand it, we get:
\[ S_{60} = 6 \left( \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \ldots + \left( \frac{1}{60} - \frac{1}{61} \right) \right) \]

### Step 6: Evaluate the telescoping sum
Most terms will cancel out, leaving us with:
\[ S_{60} = 6 \left( 1 - \frac{1}{61} \right) = 6 \left( \frac{61 - 1}{61} \right) = 6 \left( \frac{60}{61} \right) = \frac{360}{61} \]

### Final Answer
Thus, the sum of the series up to 60 terms is:
\[ \boxed{\frac{360}{61}} \]