The coefficient of `x^3` in the infinite series expansion of `(2)/((1-x) (2-x) )`, for `|x| lt 1`, is

Options

A. `-1//16`
B. `15//8`
C. `-1//8`
D. `15//16`

Answer

To find the coefficient of \( x^3 \) in the infinite series expansion of \( \frac{2}{(1-x)(2-x)} \) for \( |x| < 1 \), we can follow these steps:

### Step 1: Rewrite the Expression
We start with the expression:
\[ \frac{2}{(1-x)(2-x)} \]
We can rewrite this as:
\[ \frac{2}{(1-x)(2-x)} = \frac{2}{2(1 - \frac{x}{2})} \cdot \frac{1}{1-x} \]
This simplifies to:
\[ \frac{1}{1-x} \cdot \frac{1}{1 - \frac{x}{2}} \]

### Step 2: Use the Binomial Series Expansion
We know the binomial series expansion for \( \frac{1}{1-x} \) and \( \frac{1}{1 - \frac{x}{2}} \):
\[ \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \quad \text{and} \quad \frac{1}{1 - \frac{x}{2}} = \sum_{m=0}^{\infty} \left(\frac{x}{2}\right)^m \]
Thus, we can write:
\[ \frac{1}{1-x} = 1 + x + x^2 + x^3 + x^4 + \ldots \]
\[ \frac{1}{1 - \frac{x}{2}} = 1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} + \ldots \]

### Step 3: Multiply the Series
Now, we need to multiply these two series to find the coefficient of \( x^3 \):
\[ \left(1 + x + x^2 + x^3 + \ldots\right) \cdot \left(1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} + \ldots\right) \]

### Step 4: Identify the Coefficient of \( x^3 \)
To find the coefficient of \( x^3 \), we consider the different ways to form \( x^3 \):
1. From \( 1 \) in the first series and \( \frac{x^3}{8} \) in the second series: Coefficient = \( \frac{1}{8} \)
2. From \( x \) in the first series and \( \frac{x^2}{4} \) in the second series: Coefficient = \( 1 \cdot \frac{1}{4} = \frac{1}{4} \)
3. From \( x^2 \) in the first series and \( \frac{x}{2} \) in the second series: Coefficient = \( 1 \cdot \frac{1}{2} = \frac{1}{2} \)
4. From \( x^3 \) in the first series and \( 1 \) in the second series: Coefficient = \( 1 \cdot 1 = 1 \)

### Step 5: Sum the Coefficients
Now, we sum these coefficients:
\[ \frac{1}{8} + \frac{1}{4} + \frac{1}{2} + 1 \]
To add these, we convert them to a common denominator (which is 8):
\[ \frac{1}{8} + \frac{2}{8} + \frac{4}{8} + \frac{8}{8} = \frac{1 + 2 + 4 + 8}{8} = \frac{15}{8} \]

### Final Answer
Thus, the coefficient of \( x^3 \) in the infinite series expansion of \( \frac{2}{(1-x)(2-x)} \) is:
\[ \frac{15}{8} \]