If the expansion in power of x of the function
`(1)/(( 1 - ax)(1 - bx))` is `a_(0) + a_(1) x + a_(2) x^(2) + a_(3) x^(3) + …, ` then `a_(n)` is

Options

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

Explanation

To find the coefficient \( a_n \) in the expansion of the function

\[ \frac{1}{(1 - ax)(1 - bx)}, \]

we can follow these steps:

### Step 1: Expand Each Factor
We know that the geometric series expansion for \( \frac{1}{1 - kx} \) is given by:

\[ \frac{1}{1 - kx} = \sum_{r=0}^{\infty} k^r x^r \]

Applying this to our function, we can write:

\[ \frac{1}{1 - ax} = \sum_{r=0}^{\infty} a^r x^r \]
and
\[ \frac{1}{1 - bx} = \sum_{s=0}^{\infty} b^s x^s. \]

### Step 2: Multiply the Series
Now, we multiply these two series together:

\[ \frac{1}{(1 - ax)(1 - bx)} = \left( \sum_{r=0}^{\infty} a^r x^r \right) \left( \sum_{s=0}^{\infty} b^s x^s \right). \]

### Step 3: Find the Coefficient of \( x^n \)
The coefficient \( a_n \) of \( x^n \) in the product can be found using the Cauchy product formula. The coefficient of \( x^n \) is given by:

\[ a_n = \sum_{k=0}^{n} a^k b^{n-k}. \]

This is because we can choose \( k \) terms from the first series and \( n-k \) terms from the second series.

### Step 4: Recognize the Sum as a Finite Geometric Series
The expression \( a_n = \sum_{k=0}^{n} a^k b^{n-k} \) can be rewritten as:

\[ a_n = b^n \sum_{k=0}^{n} \left( \frac{a}{b} \right)^k. \]

This is a geometric series with \( n+1 \) terms, first term \( 1 \), and common ratio \( \frac{a}{b} \).

### Step 5: Apply the Formula for the Sum of a Geometric Series
The sum of a geometric series is given by:

\[ \sum_{k=0}^{n} r^k = \frac{1 - r^{n+1}}{1 - r} \quad \text{for } r \neq 1. \]

Applying this, we have:

\[ \sum_{k=0}^{n} \left( \frac{a}{b} \right)^k = \frac{1 - \left( \frac{a}{b} \right)^{n+1}}{1 - \frac{a}{b}} = \frac{1 - \left( \frac{a^{n+1}}{b^{n+1}} \right)}{\frac{b - a}{b}} = \frac{b}{b - a} \left( 1 - \frac{a^{n+1}}{b^{n+1}} \right). \]

### Step 6: Substitute Back into the Expression for \( a_n \)
Thus, we can substitute this back into our expression for \( a_n \):

\[ a_n = b^n \cdot \frac{b}{b - a} \left( 1 - \frac{a^{n+1}}{b^{n+1}} \right). \]

### Step 7: Simplify the Expression
This simplifies to:

\[ a_n = \frac{b^{n+1} - a^{n+1}}{b - a}. \]

### Final Result
Thus, the coefficient \( a_n \) is given by:

\[ a_n = \frac{b^{n+1} - a^{n+1}}{b - a}. \]