The value of `(1)/((1+a)(2+a))+(1)/((2+a)(3+a))+(1)/((3+a)(4+a))"......." " upto " oo` is (where, a is constant)

Options

A. A. `(1)/(1+a)`
B. B. `(2)/(1+a)`
C. C. `oo`
D. D. None of these

Answer

To solve the problem, we need to evaluate the infinite series:

\[ S = \frac{1}{(1+a)(2+a)} + \frac{1}{(2+a)(3+a)} + \frac{1}{(3+a)(4+a)} + \ldots \]

### Step 1: Rewrite the general term

The general term of the series can be expressed as:

\[ \frac{1}{(n+a)((n+1)+a)} = \frac{1}{(n+a)(n+1+a)} \]

### Step 2: Use partial fractions

We can use partial fractions to simplify the term:

\[ \frac{1}{(n+a)(n+1+a)} = \frac{A}{n+a} + \frac{B}{n+1+a} \]

To find \(A\) and \(B\), we multiply through by the denominator \((n+a)(n+1+a)\):

\[ 1 = A(n+1+a) + B(n+a) \]

Expanding this gives:

\[ 1 = An + A + Bn + Ba \]

Combining like terms:

\[ 1 = (A+B)n + (A + Ba) \]

This must hold for all \(n\), so we set up the system of equations:

1. \(A + B = 0\)
2. \(A + Ba = 1\)

From the first equation, we have \(B = -A\). Substituting into the second equation gives:

\[ A - Aa = 1 \implies A(1-a) = 1 \implies A = \frac{1}{1-a} \]

Thus, \(B = -\frac{1}{1-a}\).

### Step 3: Rewrite the series using partial fractions

Now we can rewrite the general term:

\[ \frac{1}{(n+a)(n+1+a)} = \frac{1}{1-a} \left( \frac{1}{n+a} - \frac{1}{n+1+a} \right) \]

### Step 4: Write the series

Substituting this back into the series gives:

\[ S = \frac{1}{1-a} \left( \left( \frac{1}{1+a} - \frac{1}{2+a} \right) + \left( \frac{1}{2+a} - \frac{1}{3+a} \right) + \left( \frac{1}{3+a} - \frac{1}{4+a} \right) + \ldots \right) \]

### Step 5: Recognize the telescoping nature

Notice that this is a telescoping series. Most terms will cancel out:

\[ S = \frac{1}{1-a} \left( \frac{1}{1+a} - \lim_{n \to \infty} \frac{1}{n+1+a} \right) \]

As \(n \to \infty\), \(\frac{1}{n+1+a} \to 0\). Thus, we have:

\[ S = \frac{1}{1-a} \cdot \frac{1}{1+a} \]

### Step 6: Final expression

Combining the fractions gives:

\[ S = \frac{1}{(1+a)(1-a)} = \frac{1}{1-a^2} \]

### Conclusion

Thus, the value of the infinite series is:

\[ S = \frac{1}{1+a} \]