If `|a| lt 1 ` and |b|` lt` 1 , then the sum of the series
`a(a+b) + a^2 (a^2 + b^2) + a^3 (a^3 + b^3) `+… upto `oo` is :

Options

A. a. `(a^2)/(b) +1`
B. b. `(a^2)/(1-ab)`
C. c. `(a^2)/(1-a^2) + (ab)/(1-ab)`
D. d. `(a^2)/(b^2) + (1)/(ab)`

Answer

To find the sum of the series \( S = a(a+b) + a^2(a^2+b^2) + a^3(a^3+b^3) + \ldots \) up to infinity, we can break it down step by step.

### Step 1: Identify the structure of the series
The series can be rewritten as:
\[ S = a(a+b) + a^2(a^2+b^2) + a^3(a^3+b^3) + \ldots \]
This can be expressed as:
\[ S = a^2 + ab + a^4 + a^2b^2 + a^6 + a^3b^3 + \ldots \]
We can separate the terms involving \( a \) and \( b \).

### Step 2: Group the terms
We can group the terms based on powers of \( a \) and \( b \):
\[ S = (a^2 + a^4 + a^6 + \ldots) + (ab + a^2b^2 + a^3b^3 + \ldots) \]

### Step 3: Recognize the geometric series
The first group \( a^2 + a^4 + a^6 + \ldots \) is a geometric series with the first term \( a^2 \) and common ratio \( a^2 \):
\[ \text{Sum of first series} = \frac{a^2}{1 - a^2} \quad \text{(since } |a| < 1\text{)} \]

The second group \( ab + a^2b^2 + a^3b^3 + \ldots \) is also a geometric series with the first term \( ab \) and common ratio \( ab \):
\[ \text{Sum of second series} = \frac{ab}{1 - ab} \quad \text{(since } |ab| < 1\text{)} \]

### Step 4: Combine the results
Now, we can combine the sums of both series:
\[ S = \frac{a^2}{1 - a^2} + \frac{ab}{1 - ab} \]

### Step 5: Simplify the expression
To express \( S \) in a single fraction:
\[ S = \frac{a^2(1 - ab) + ab(1 - a^2)}{(1 - a^2)(1 - ab)} \]
Expanding the numerator:
\[ S = \frac{a^2 - a^2ab + ab - a^2b}{(1 - a^2)(1 - ab)} \]
Combining like terms:
\[ S = \frac{a^2 + ab - a^2ab - a^2b}{(1 - a^2)(1 - ab)} \]

### Final Result
Thus, the sum of the series is:
\[ S = \frac{a^2 + ab}{(1 - a^2)(1 - ab)} \]