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 terms of the sequence given by \( t_r = \frac{1}{(r+1)(r+3)} \), we can follow these steps:

### Step 1: Rewrite the term
We start with the term:
\[ t_r = \frac{1}{(r+1)(r+3)} \]
We can use partial fraction decomposition to rewrite this term:
\[ t_r = \frac{A}{r+1} + \frac{B}{r+3} \]
Multiplying through by the denominator \((r+1)(r+3)\) gives:
\[ 1 = A(r+3) + B(r+1) \]
Expanding this, we have:
\[ 1 = Ar + 3A + Br + B \]
Combining like terms, we get:
\[ 1 = (A + B)r + (3A + B) \]
From this, we can set up the equations:
1. \( A + B = 0 \)
2. \( 3A + B = 1 \)

### Step 2: Solve for A and B
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 the term using A and B
Now substituting \( A \) and \( B \) back into the partial fractions:
\[ t_r = \frac{1/2}{r+1} - \frac{1/2}{r+3} \]
Thus, we can write:
\[ t_r = \frac{1}{2} \left( \frac{1}{r+1} - \frac{1}{r+3} \right) \]

### Step 4: Sum the series
Now we need to find the sum of the infinite series:
\[ S = \sum_{r=1}^{\infty} t_r = \sum_{r=1}^{\infty} \frac{1}{2} \left( \frac{1}{r+1} - \frac{1}{r+3} \right) \]
This is a telescoping series. Writing out the first few terms:
\[ S = \frac{1}{2} \left( \left( \frac{1}{2} - \frac{1}{4} \right) + \left( \frac{1}{3} - \frac{1}{5} \right) + \left( \frac{1}{4} - \frac{1}{6} \right) + \ldots \right) \]

### Step 5: Identify the cancellation
Notice that most terms will cancel:
\[ S = \frac{1}{2} \left( \frac{1}{2} + \frac{1}{3} - \lim_{n \to \infty} \left( \frac{1}{n+2} + \frac{1}{n+3} \right) \right) \]
As \( n \to \infty \), \( \frac{1}{n+2} \) and \( \frac{1}{n+3} \) approach 0. Therefore:
\[ S = \frac{1}{2} \left( \frac{1}{2} + \frac{1}{3} \right) = \frac{1}{2} \left( \frac{3}{6} + \frac{2}{6} \right) = \frac{1}{2} \cdot \frac{5}{6} = \frac{5}{12} \]

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