The sum of the infinite series `1 + (1+a) x + (1 + a + a^(2)) x^(2) + (1 + a + a^(2) + a^(3)) x^(3)+…` where `0 lt a, x lt 1`, is

Options

A. `1//(1 - x) (1 - a)`
B. `1//(1-x)(1-ax)`
C. `1//(1-a) (1 - ax)`
D. `1//(1-x) (1+a)`

Answer

To find the sum of the infinite series

\[ S = 1 + (1 + a)x + (1 + a + a^2)x^2 + (1 + a + a^2 + a^3)x^3 + \ldots \]

where \(0 < a, x < 1\), we can follow these steps:

### Step 1: Rewrite the series
We can express the series in a more manageable form. Notice that the \(n\)-th term of the series can be expressed as:

\[ T_n = (1 + a + a^2 + \ldots + a^n)x^n \]

So, we can rewrite the series as:

\[ S = \sum_{n=0}^{\infty} (1 + a + a^2 + \ldots + a^n)x^n \]

### Step 2: Use the formula for the sum of a geometric series
The sum of the first \(n\) terms of a geometric series can be expressed as:

\[ 1 + a + a^2 + \ldots + a^n = \frac{1 - a^{n+1}}{1 - a} \]

Thus, we can rewrite \(T_n\):

\[ T_n = \frac{1 - a^{n+1}}{1 - a} x^n \]

### Step 3: Substitute back into the series
Now we can substitute \(T_n\) back into the series \(S\):

\[ S = \sum_{n=0}^{\infty} \frac{1 - a^{n+1}}{1 - a} x^n \]

### Step 4: Split the series
We can split the series into two parts:

\[ S = \frac{1}{1 - a} \sum_{n=0}^{\infty} x^n - \frac{1}{1 - a} \sum_{n=0}^{\infty} a^{n+1} x^n \]

### Step 5: Calculate the first series
The first series is a standard geometric series:

\[ \sum_{n=0}^{\infty} x^n = \frac{1}{1 - x} \]

### Step 6: Calculate the second series
The second series can be factored:

\[ \sum_{n=0}^{\infty} a^{n+1} x^n = a \sum_{n=0}^{\infty} (ax)^n = \frac{a}{1 - ax} \]

### Step 7: Combine the results
Now we can combine the results:

\[ S = \frac{1}{1 - a} \left( \frac{1}{1 - x} - \frac{a}{1 - ax} \right) \]

### Step 8: Simplify the expression
To simplify, we can find a common denominator:

\[ S = \frac{1}{1 - a} \cdot \frac{(1 - ax) - a(1 - x)}{(1 - x)(1 - ax)} \]

This simplifies to:

\[ S = \frac{1 - ax - a + ax}{(1 - a)(1 - x)(1 - ax)} = \frac{1 - a}{(1 - a)(1 - x)(1 - ax)} = \frac{1}{(1 - x)(1 - ax)} \]

### Final Result
Thus, the sum of the infinite series is:

\[ S = \frac{1}{(1 - x)(1 - ax)} \]