The coefficient of `x^(n)` in the expansion of `(1)/((1-ax)(1-bx))` is :

Options

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

Answer

To find the coefficient of \( x^n \) in the expansion of \( \frac{1}{(1 - ax)(1 - bx)} \), we can use the method of partial fractions and the binomial series expansion. Here’s a step-by-step solution:

### Step 1: Rewrite the expression
We start with the expression:
\[ \frac{1}{(1 - ax)(1 - bx)} \]
This can be rewritten using partial fractions:
\[ \frac{1}{(1 - ax)(1 - bx)} = \frac{A}{1 - ax} + \frac{B}{1 - bx} \]
where \( A \) and \( B \) are constants to be determined.

### Step 2: Find constants A and B
Multiplying through by \( (1 - ax)(1 - bx) \) gives:
\[ 1 = A(1 - bx) + B(1 - ax) \]
Expanding this, we have:
\[ 1 = A - Abx + B - Bax \]
Grouping like terms, we get:
\[ 1 = (A + B) - (Ab + Ba)x \]
For this to hold for all \( x \), we need:
1. \( A + B = 1 \)
2. \( -Ab - Ba = 0 \)

From the second equation, we can express \( B \) in terms of \( A \):
\[ B = \frac{Ab}{a} \]
Substituting \( B \) in the first equation:
\[ A + \frac{Ab}{a} = 1 \]
Factoring out \( A \):
\[ A(1 + \frac{b}{a}) = 1 \implies A = \frac{1}{1 + \frac{b}{a}} = \frac{a}{a + b} \]
Now substituting back to find \( B \):
\[ B = 1 - A = 1 - \frac{a}{a + b} = \frac{b}{a + b} \]

### Step 3: Expand using the binomial series
Now we can write:
\[ \frac{1}{(1 - ax)(1 - bx)} = \frac{a}{a + b} \cdot \frac{1}{1 - ax} + \frac{b}{a + b} \cdot \frac{1}{1 - bx} \]
Using the binomial series expansion:
\[ \frac{1}{1 - kx} = \sum_{n=0}^{\infty} k^n x^n \]
we have:
\[ \frac{1}{1 - ax} = \sum_{n=0}^{\infty} (ax)^n = \sum_{n=0}^{\infty} a^n x^n \]
and
\[ \frac{1}{1 - bx} = \sum_{n=0}^{\infty} (bx)^n = \sum_{n=0}^{\infty} b^n x^n \]

### Step 4: Combine the expansions
Now substituting back, we get:
\[ \frac{1}{(1 - ax)(1 - bx)} = \frac{a}{a + b} \sum_{n=0}^{\infty} a^n x^n + \frac{b}{a + b} \sum_{n=0}^{\infty} b^n x^n \]
To find the coefficient of \( x^n \), we look at the terms:
\[ \frac{a}{a + b} a^n + \frac{b}{a + b} b^n \]

### Step 5: Final expression
Thus, the coefficient of \( x^n \) in the expansion is:
\[ \frac{a^{n+1}}{a + b} + \frac{b^{n+1}}{a + b} \]
This can be simplified to:
\[ \frac{a^{n+1} + b^{n+1}}{a + b} \]

### Final Result
The coefficient of \( x^n \) in the expansion of \( \frac{1}{(1 - ax)(1 - bx)} \) is:
\[ \frac{a^{n+1} + b^{n+1}}{a + b} \]