If sum of x terms of a series is `S_(x)=(1)/((2x+3)(2x+1))`
whose `r^(th)` term is `T_(r)`. Then, `sum_(r=1)^(n) (1)/(T_(r))` is equal to

Options

A. `(1)/(4)sum(2r+1)(2r-1)(2r+3)`
B. `-(1)/(4)sum(2r+1)(2r-1)(2r+3)`
C. `sum(2r+1)(2r-1)(2r+3)`
D. none of these

Explanation

To solve the problem, we need to find the sum \( \sum_{r=1}^{n} \frac{1}{T_r} \) where \( T_r \) is the \( r^{th} \) term of the series defined by the sum of \( x \) terms \( S_x = \frac{1}{(2x+3)(2x+1)} \).

### Step 1: Find the \( r^{th} \) term \( T_r \)

We know that the \( r^{th} \) term \( T_r \) can be expressed as:

\[ T_r = S_r - S_{r-1} \]

Substituting the given expression for \( S_x \):

\[ S_r = \frac{1}{(2r+3)(2r+1)} \quad \text{and} \quad S_{r-1} = \frac{1}{(2(r-1)+3)(2(r-1)+1)} = \frac{1}{(2r+1)(2r-1)} \]

Now, we can calculate \( T_r \):

\[ T_r = \frac{1}{(2r+3)(2r+1)} - \frac{1}{(2r+1)(2r-1)} \]

### Step 2: Simplify \( T_r \)

To simplify \( T_r \), we need a common denominator, which is \( (2r+3)(2r+1)(2r-1) \):

\[ T_r = \frac{(2r-1) - (2r+3)}{(2r+3)(2r+1)(2r-1)} \]

Calculating the numerator:

\[ (2r-1) - (2r+3) = 2r - 1 - 2r - 3 = -4 \]

Thus, we have:

\[ T_r = \frac{-4}{(2r+3)(2r+1)(2r-1)} \]

### Step 3: Find \( \frac{1}{T_r} \)

Now, we need to find \( \frac{1}{T_r} \):

\[ \frac{1}{T_r} = \frac{(2r+3)(2r+1)(2r-1)}{-4} \]

### Step 4: Calculate the summation \( \sum_{r=1}^{n} \frac{1}{T_r} \)

Now we can write the summation:

\[ \sum_{r=1}^{n} \frac{1}{T_r} = \sum_{r=1}^{n} \frac{(2r+3)(2r+1)(2r-1)}{-4} \]

This can be simplified to:

\[ = -\frac{1}{4} \sum_{r=1}^{n} (2r+3)(2r+1)(2r-1) \]

### Step 5: Final Result

The final result is:

\[ \sum_{r=1}^{n} \frac{1}{T_r} = -\frac{1}{4} \sum_{r=1}^{n} (2r+3)(2r+1)(2r-1) \]