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

Options

A. `(n^(3))/(2(n+1)(n+2)(n+3))`
B. `(n^(3)+6n^(2)-3n)/(6(n+2)(n+3)(n+4))`
C. `(15n^(2)+7n)/(4n(n+1)(n+5))`
D. `(n^(3)+6n^(2)+11n)/((18n+1)(n+2)(n+3))`

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 denote the k-th term of the series as:

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

### Step 1: Rewrite the k-th term
We can simplify \(u_k\) using partial fractions. We can express:

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

### Step 2: Find the coefficients
To find A, B, C, and D, we multiply through by the denominator \(k(k+1)(k+2)(k+3)\) and set up the equation:

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

Expanding this and equating coefficients will give us a system of equations to solve for A, B, C, and D.

### Step 3: Calculate the sum
Once we have the partial fraction decomposition, we can express \(S_n\) as:

\[ S_n = \sum_{k=1}^{n} u_k = \sum_{k=1}^{n} \left( \frac{A}{k} + \frac{B}{k+1} + \frac{C}{k+2} + \frac{D}{k+3} \right) \]

This can be separated into four distinct sums:

\[ S_n = A \sum_{k=1}^{n} \frac{1}{k} + B \sum_{k=1}^{n} \frac{1}{k+1} + C \sum_{k=1}^{n} \frac{1}{k+2} + D \sum_{k=1}^{n} \frac{1}{k+3} \]

### Step 4: Evaluate the sums
Using the properties of harmonic sums, we can evaluate these sums. The harmonic sum \(H_n\) is defined as:

\[ H_n = \sum_{k=1}^{n} \frac{1}{k} \]

Thus, we can express \(S_n\) in terms of \(H_n\):

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

### Step 5: Final simplification
After substituting the values of A, B, C, and D, we can combine like terms to find a simplified expression for \(S_n\).

### Conclusion
The final expression for the sum of the first n terms of the series will be:

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

This gives us the sum of the first n terms of the series.