Find the coefficient of `x^(3)` in the power series expansion of `(5x+6)/((x+2)(1-x))` specifying the region in which the expansion is valid.

Answer

To find the coefficient of \( x^3 \) in the power series expansion of \( \frac{5x + 6}{(x + 2)(1 - x)} \), we will follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ \frac{5x + 6}{(x + 2)(1 - x)} \]
We can express this as:
\[ (5x + 6) \cdot \frac{1}{(x + 2)(1 - x)} \]

### Step 2: Expand the denominator
We can expand \( \frac{1}{(x + 2)(1 - x)} \) using partial fractions. We first rewrite it as:
\[ \frac{1}{(x + 2)(1 - x)} = \frac{A}{x + 2} + \frac{B}{1 - x} \]
Multiplying through by the denominator \( (x + 2)(1 - x) \) gives:
\[ 1 = A(1 - x) + B(x + 2) \]
Expanding this, we have:
\[ 1 = A - Ax + Bx + 2B \]
Grouping terms:
\[ 1 = (A + B)x + (A + 2B) \]
Setting coefficients equal, we get the system:
1. \( A + B = 0 \)
2. \( A + 2B = 1 \)

### Step 3: Solve for A and B
From the first equation, \( A = -B \). Substituting into the second equation:
\[ -B + 2B = 1 \implies B = 1 \implies A = -1 \]
Thus, we have:
\[ \frac{1}{(x + 2)(1 - x)} = \frac{-1}{x + 2} + \frac{1}{1 - x} \]

### Step 4: Expand each term
Now we can expand each term:
1. For \( \frac{-1}{x + 2} \):
\[ \frac{-1}{x + 2} = -\frac{1}{2} \cdot \frac{1}{1 - (-\frac{x}{2})} = -\frac{1}{2} \sum_{n=0}^{\infty} \left(-\frac{x}{2}\right)^n = -\frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{2^n} \]
This series converges for \( |x| < 2 \).

2. For \( \frac{1}{1 - x} \):
\[ \frac{1}{1 - x} = \sum_{n=0}^{\infty} x^n \]
This series converges for \( |x| < 1 \).

### Step 5: Combine the series
Combining both expansions:
\[ \frac{5x + 6}{(x + 2)(1 - x)} = (5x + 6) \left(-\frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{2^n} + \sum_{n=0}^{\infty} x^n\right) \]

### Step 6: Find the coefficient of \( x^3 \)
We need to find the coefficient of \( x^3 \):
1. From \( 5x \cdot \sum_{n=0}^{\infty} x^n \): The coefficient of \( x^2 \) is \( 5 \).
2. From \( 6 \cdot \sum_{n=0}^{\infty} x^n \): The coefficient of \( x^3 \) is \( 6 \).
3. From \( 5x \cdot \left(-\frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{2^n}\right) \): The coefficient of \( x^2 \) is \( -\frac{5}{2} \).
4. From \( 6 \cdot \left(-\frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{2^n}\right) \): The coefficient of \( x^3 \) is \( -\frac{6}{2} = -3 \).

### Step 7: Combine the coefficients
Now, we combine the coefficients:
\[ \text{Coefficient of } x^3 = 6 - 3 + \frac{5}{4} - \frac{5}{2} \]
Calculating this gives:
\[ = 6 - 3 + \frac{5}{4} - \frac{10}{4} = 3 - \frac{5}{4} = \frac{12}{4} - \frac{5}{4} = \frac{7}{4} \]

### Final Answer
The coefficient of \( x^3 \) in the power series expansion is \( \frac{7}{4} \).

### Region of Validity
The series converges for \( |x| < 1 \) (from \( \frac{1}{1 - x} \)) and \( |x| < 2 \) (from \( \frac{-1}{x + 2} \)). Therefore, the region of validity is:
\[ |x| < 1 \]