The sum to 50 terms of the series
`3/1^2+5/(1^2+2^2)+7/(1^+2^2+3^2)+….+… is `

Options

A. `100/17`
B. `150/17`
C. `200/51`
D. `50/17`

Explanation

To find the sum to 50 terms of the series

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

we can break it down step by step.

### Step 1: Identify the Numerator
The numerators of the series are: 3, 5, 7, ...

This is an arithmetic progression (AP) where:
- First term \( a = 3 \)
- Common difference \( d = 2 \)

The \( n \)-th term of this AP can be expressed as:
\[ a_n = a + (n - 1)d = 3 + (n - 1) \cdot 2 = 2n + 1 \]

### Step 2: Identify the Denominator
The denominators of the series are the sums of squares of the first \( n \) natural numbers. The formula for the sum of squares of the first \( n \) natural numbers is:
\[ S_n = \frac{n(n + 1)(2n + 1)}{6} \]

Thus, the \( n \)-th term of the series can be expressed as:
\[ T_n = \frac{2n + 1}{S_n} = \frac{2n + 1}{\frac{n(n + 1)(2n + 1)}{6}} = \frac{6(2n + 1)}{n(n + 1)(2n + 1)} \]

### Step 3: Simplify the \( n \)-th Term
We can simplify \( T_n \):
\[ T_n = \frac{6}{n(n + 1)} \]

### Step 4: Rewrite the Series
Now we need to find the sum of the first 50 terms:
\[ S = \sum_{n=1}^{50} T_n = \sum_{n=1}^{50} \frac{6}{n(n + 1)} \]

### Step 5: Use Partial Fraction Decomposition
We can express \( \frac{6}{n(n + 1)} \) as:
\[ \frac{6}{n(n + 1)} = 6 \left( \frac{1}{n} - \frac{1}{n + 1} \right) \]

### Step 6: Write the Series in Summation Form
Thus, the sum becomes:
\[ S = 6 \sum_{n=1}^{50} \left( \frac{1}{n} - \frac{1}{n + 1} \right) \]

### Step 7: Evaluate the Series
This is a telescoping series:
\[ S = 6 \left( \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \ldots + \left( \frac{1}{50} - \frac{1}{51} \right) \right) \]

Most terms cancel out, leaving:
\[ S = 6 \left( 1 - \frac{1}{51} \right) = 6 \cdot \frac{50}{51} = \frac{300}{51} \]

### Step 8: Simplify the Result
Simplifying \( \frac{300}{51} \):
\[ \frac{300}{51} = \frac{100}{17} \]

Thus, the sum of the series up to 50 terms is:
\[ \boxed{\frac{100}{17}} \]