The sum of infinite terms of the sequence whose `r^("th")` term is given by `t_(r)=(1)/((r+1)(r+3))` is equal to

Options

A. 1
B. 2
C. `(3)/(4)`
D. `(5)/(12)`

Answer

To find the sum of the infinite series given by the term \( t_r = \frac{1}{(r+1)(r+3)} \), we can follow these steps:

### Step 1: Rewrite the term using partial fractions
We start with the term:
\[ t_r = \frac{1}{(r+1)(r+3)} \]
We can express this using partial fractions:
\[ t_r = \frac{A}{r+1} + \frac{B}{r+3} \]
Multiplying both sides by \((r+1)(r+3)\) gives:
\[ 1 = A(r+3) + B(r+1) \]
Expanding this, we have:
\[ 1 = Ar + 3A + Br + B \]
Combining like terms:
\[ 1 = (A + B)r + (3A + B) \]
Setting coefficients equal, we get the system:
1. \( A + B = 0 \)
2. \( 3A + B = 1 \)

### Step 2: Solve the system of equations
From the first equation, we can express \( B \) in terms of \( A \):
\[ B = -A \]
Substituting into the second equation:
\[ 3A - A = 1 \implies 2A = 1 \implies A = \frac{1}{2} \]
Then substituting back to find \( B \):
\[ B = -\frac{1}{2} \]

### Step 3: Rewrite \( t_r \)
Now we can rewrite \( t_r \):
\[ t_r = \frac{1/2}{r+1} - \frac{1/2}{r+3} \]
Thus,
\[ t_r = \frac{1}{2} \left( \frac{1}{r+1} - \frac{1}{r+3} \right) \]

### Step 4: Set up the series
Now we can sum the series:
\[ S = \sum_{r=1}^{\infty} t_r = \sum_{r=1}^{\infty} \frac{1}{2} \left( \frac{1}{r+1} - \frac{1}{r+3} \right) \]
Factoring out the constant:
\[ S = \frac{1}{2} \sum_{r=1}^{\infty} \left( \frac{1}{r+1} - \frac{1}{r+3} \right) \]

### Step 5: Recognize the telescoping series
The series is telescoping. Writing out the first few terms:
\[ S = \frac{1}{2} \left( \left( \frac{1}{2} - \frac{1}{4} \right) + \left( \frac{1}{3} - \frac{1}{5} \right) + \left( \frac{1}{4} - \frac{1}{6} \right) + \ldots \right) \]
Most terms will cancel out, leaving:
\[ S = \frac{1}{2} \left( \frac{1}{2} + \frac{1}{3} \right) \]

### Step 6: Calculate the remaining terms
Calculating the remaining terms:
\[ S = \frac{1}{2} \left( \frac{3}{6} + \frac{2}{6} \right) = \frac{1}{2} \cdot \frac{5}{6} = \frac{5}{12} \]

### Final Answer
Thus, the sum of the infinite series is:
\[ \boxed{\frac{5}{12}} \]