If the sum of n term of the series `(5)/(1.2.3) + (6)/(2.3.4) + ( 7)/(3.4.5)+"…."` is `a/2 - (n+b)/((n+1)(n+2))`, then

Options

A. `a = 3`
B. `b=2`
C. `b =3`
D. `a=2`

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) in the expression for the sum of the first \( n \) terms of the series given by:

\[ S_n = \frac{5}{1 \cdot 2 \cdot 3} + \frac{6}{2 \cdot 3 \cdot 4} + \frac{7}{3 \cdot 4 \cdot 5} + \ldots \]

The sum is expressed in the form:

\[ S_n = \frac{a}{2} - \frac{n + b}{(n + 1)(n + 2)} \]

### Step 1: Identify the nth term of the series

The general term of the series can be expressed as:

\[ T_n = \frac{n + 4}{n(n + 1)(n + 2)} \]

This is derived from observing the pattern in the numerators and denominators of the series.

### Step 2: Write the sum of the first n terms

The sum of the first \( n \) terms can be expressed as:

\[ S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \frac{k + 4}{k(k + 1)(k + 2)} \]

### Step 3: Decompose the term into partial fractions

We can decompose \( \frac{k + 4}{k(k + 1)(k + 2)} \) into partial fractions:

\[ \frac{k + 4}{k(k + 1)(k + 2)} = \frac{A}{k} + \frac{B}{k + 1} + \frac{C}{k + 2} \]

Multiplying through by the denominator \( k(k + 1)(k + 2) \) and equating coefficients will help us find \( A \), \( B \), and \( C \).

### Step 4: Solve for A, B, and C

Expanding and comparing coefficients, we find:

1. \( A + B + C = 0 \)
2. \( 3A + 2B + C = 1 \)
3. \( 2A + B = 4 \)

From these equations, we can solve for \( A \), \( B \), and \( C \).

### Step 5: Calculate the values

After solving the equations, we find:

- \( A = 2 \)
- \( B = -3 \)
- \( C = 1 \)

Thus, we can rewrite the sum:

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

### Step 6: Evaluate the summation

Now we can evaluate the summation term by term:

1. The sum of \( \frac{2}{k} \) gives \( 2 \sum_{k=1}^{n} \frac{1}{k} \).
2. The sum of \( -\frac{3}{k + 1} \) gives \( -3 \sum_{k=2}^{n+1} \frac{1}{k} \).
3. The sum of \( \frac{1}{k + 2} \) gives \( \sum_{k=3}^{n+2} \frac{1}{k} \).

### Step 7: Combine results

Combining these results leads to:

\[ S_n = 2H_n - 3H_{n+1} + H_{n+2} \]

Where \( H_n \) is the \( n \)-th harmonic number.

### Step 8: Simplify and compare

After simplification, we can express \( S_n \) in the required form and compare coefficients to find \( a \) and \( b \).

### Conclusion

After all calculations, we find:

- \( a = 3 \)
- \( b = 3 \)

Thus, the final answer is:

\[ \boxed{(a, b) = (3, 3)} \]