If `|a| < 1 and |b| < 1,` then the sum of the series `a(a+b)+a^2(a^2+b^2)+a^3(a^3+b^3)+.....oo` is

Options

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

Answer

To find the sum of the series \( S = a(a+b) + a^2(a^2+b^2) + a^3(a^3+b^3) + \ldots \) where \( |a| < 1 \) and \( |b| < 1 \), we can follow these steps:

### Step 1: Identify the General Term
The general term of the series can be expressed as:
\[ T_r = a^r (a^r + b^r) \]
This can be rewritten as:
\[ T_r = a^r \cdot a^r + a^r \cdot b^r = a^{2r} + a^r b^r \]

### Step 2: Write the Series
Now we can express the series \( S \) as:
\[ S = \sum_{r=1}^{\infty} (a^{2r} + a^r b^r) \]
This can be separated into two series:
\[ S = \sum_{r=1}^{\infty} a^{2r} + \sum_{r=1}^{\infty} a^r b^r \]

### Step 3: Sum the First Series
The first series \( \sum_{r=1}^{\infty} a^{2r} \) is a geometric series with first term \( a^2 \) and common ratio \( a^2 \):
\[ \sum_{r=1}^{\infty} a^{2r} = \frac{a^2}{1 - a^2} \quad \text{(since } |a| < 1\text{)} \]

### Step 4: Sum the Second Series
The second series \( \sum_{r=1}^{\infty} a^r b^r \) can also be written as a geometric series with first term \( ab \) and common ratio \( ab \):
\[ \sum_{r=1}^{\infty} a^r b^r = \frac{ab}{1 - ab} \quad \text{(since } |ab| < 1\text{)} \]

### Step 5: Combine the Results
Now we can combine both results to find the total sum \( S \):
\[ S = \frac{a^2}{1 - a^2} + \frac{ab}{1 - ab} \]

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