Sum of n terms of the series `1/(1.2.3.4.)+1/(2.3.4.5) +1/(3.4.5.6)+....`

Options

A. `1/6log_(e)2-(1)/(24)`
B. `5/2-log_(e)2`
C. `3/2-log_(e)2`
D. none of these

Answer

To find the sum of the first \( n \) terms of the series

\[ S_n = \frac{1}{1 \cdot 2 \cdot 3 \cdot 4} + \frac{1}{2 \cdot 3 \cdot 4 \cdot 5} + \frac{1}{3 \cdot 4 \cdot 5 \cdot 6} + \ldots \]

we can express the general term of the series as:

\[ T_k = \frac{1}{k(k+1)(k+2)(k+3)} \]

### Step 1: Rewrite the General Term
We can rewrite \( T_k \) by using partial fractions. We want to express:

\[ \frac{1}{k(k+1)(k+2)(k+3)} = \frac{A}{k} + \frac{B}{k+1} + \frac{C}{k+2} + \frac{D}{k+3} \]

Multiplying through by the denominator \( k(k+1)(k+2)(k+3) \) gives:

\[ 1 = A(k+1)(k+2)(k+3) + B(k)(k+2)(k+3) + C(k)(k+1)(k+3) + D(k)(k+1)(k+2) \]

### Step 2: Solve for Coefficients
To find \( A, B, C, D \), we can substitute suitable values for \( k \) to simplify the calculations.

1. Let \( k = 0 \):
\[ 1 = A(1)(2)(3) \implies A = \frac{1}{6} \]

2. Let \( k = -1 \):
\[ 1 = B(-1)(1)(2) \implies B = -\frac{1}{2} \]

3. Let \( k = -2 \):
\[ 1 = C(-2)(-1)(1) \implies C = \frac{1}{2} \]

4. Let \( k = -3 \):
\[ 1 = D(-3)(-2)(-1) \implies D = -\frac{1}{6} \]

Thus, we have:

\[ \frac{1}{k(k+1)(k+2)(k+3)} = \frac{1}{6k} - \frac{1}{2(k+1)} + \frac{1}{2(k+2)} - \frac{1}{6(k+3)} \]

### Step 3: Write the Sum
Now we can express \( S_n \) as:

\[ S_n = \sum_{k=1}^{n} \left( \frac{1}{6k} - \frac{1}{2(k+1)} + \frac{1}{2(k+2)} - \frac{1}{6(k+3)} \right) \]

### Step 4: Simplify the Series
This series can be simplified by observing the telescoping nature of the terms.

- The \( -\frac{1}{2(k+1)} \) and \( \frac{1}{2(k+2)} \) terms will cancel out with subsequent terms.
- The remaining terms will include contributions from the first few and the last few terms.

### Step 5: Calculate the Result
After performing the summation and simplifying, we can find that:

\[ S_n = \frac{1}{6} \left( 1 - \frac{1}{(n+1)(n+2)(n+3)} \right) \]

### Final Result
Thus, the sum of the first \( n \) terms of the series is:

\[ S_n = \frac{1}{6} \left( 1 - \frac{1}{(n+1)(n+2)(n+3)} \right) \]