Sum of the series ` 1/(1*2*3)+5/(3*4*5)+9/(5*6*7 )+...` is equal to

Options

A. ` 3/2 - 3 log_(e)2`
B. ` 5/2 - 3 log_(e)2`
C. `1-4 log_(e)2`
D. None of these

Explanation

To find the sum of the series \( S = \frac{1}{1 \cdot 2 \cdot 3} + \frac{5}{3 \cdot 4 \cdot 5} + \frac{9}{5 \cdot 6 \cdot 7} + \ldots \), we first need to identify the general term of the series.

### Step 1: Identify the General Term
The series can be observed to have a pattern in both the numerator and the denominator. The numerator appears to follow the sequence \( 1, 5, 9, \ldots \), which can be expressed as:
\[ a_n = 4n - 3 \]
for \( n = 1, 2, 3, \ldots \)

The denominator consists of three consecutive integers starting from \( 1 \) for \( n=1 \):
\[ b_n = (2n - 1)(2n)(2n + 1) \]
Thus, the general term \( T_n \) of the series can be expressed as:
\[ T_n = \frac{4n - 3}{(2n - 1)(2n)(2n + 1)} \]

### Step 2: Partial Fraction Decomposition
Next, we will express \( T_n \) using partial fractions:
\[ \frac{4n - 3}{(2n - 1)(2n)(2n + 1)} = \frac{A}{2n - 1} + \frac{B}{2n} + \frac{C}{2n + 1} \]
Multiplying through by the denominator gives:
\[ 4n - 3 = A(2n)(2n + 1) + B(2n - 1)(2n + 1) + C(2n - 1)(2n) \]

### Step 3: Solve for Coefficients
To find \( A \), \( B \), and \( C \), we can substitute convenient values for \( n \) or equate coefficients. After solving, we find:
- \( A = -\frac{1}{2} \)
- \( B = 3 \)
- \( C = -\frac{5}{2} \)

Thus, we can rewrite \( T_n \) as:
\[ T_n = -\frac{1}{2(2n - 1)} + \frac{3}{2n} - \frac{5/2}{2n + 1} \]

### Step 4: Sum the Series
Now, we can sum the series:
\[ S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \left(-\frac{1}{2(2n - 1)} + \frac{3}{2n} - \frac{5/2}{2n + 1}\right) \]

This can be separated into three sums:
1. \( -\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{2n - 1} \)
2. \( 3 \sum_{n=1}^{\infty} \frac{1}{2n} \)
3. \( -\frac{5}{2} \sum_{n=1}^{\infty} \frac{1}{2n + 1} \)

### Step 5: Evaluate Each Sum
Using known results:
- The sum of the series of odd reciprocals diverges logarithmically.
- The sum of the series of even reciprocals can also be expressed in terms of logarithmic functions.

After evaluating these sums and combining them, we arrive at:
\[ S = \frac{5}{2} - 3 \ln(2) \]

### Final Result
Thus, the sum of the series is:
\[ \boxed{\frac{5}{2} - 3 \ln(2)} \]