Find the coefficient of `x^(4)` in the expansion of `(3x)/((x-2)(x+1))` in powers of x specifying the interval in which the expansion is valid.

Answer

To find the coefficient of \( x^4 \) in the expansion of \( \frac{3x}{(x-2)(x+1)} \), we will use partial fraction decomposition and then apply the binomial theorem for expansion. Let's go through the steps systematically.

### Step 1: Partial Fraction Decomposition
We start with the expression:
\[ \frac{3x}{(x-2)(x+1)} \]
We can express this as:
\[ \frac{3x}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1} \]
where \( A \) and \( B \) are constants we need to determine.

### Step 2: Finding Constants \( A \) and \( B \)
To find \( A \) and \( B \), we multiply both sides by \( (x-2)(x+1) \):
\[ 3x = A(x+1) + B(x-2) \]
Expanding the right side gives:
\[ 3x = Ax + A + Bx - 2B \]
Combining like terms:
\[ 3x = (A + B)x + (A - 2B) \]
This leads to the system of equations:
1. \( A + B = 3 \)
2. \( A - 2B = 0 \)

### Step 3: Solving the System of Equations
From the second equation, we can express \( A \) in terms of \( B \):
\[ A = 2B \]
Substituting into the first equation:
\[ 2B + B = 3 \implies 3B = 3 \implies B = 1 \]
Now substituting \( B = 1 \) back into \( A = 2B \):
\[ A = 2 \times 1 = 2 \]
Thus, we have:
\[ A = 2, \quad B = 1 \]

### Step 4: Writing the Partial Fraction Decomposition
Now we can rewrite the original expression:
\[ \frac{3x}{(x-2)(x+1)} = \frac{2}{x-2} + \frac{1}{x+1} \]

### Step 5: Expanding Using Binomial Theorem
We can rewrite the fractions:
\[ \frac{2}{x-2} = -\frac{2}{2-x} = -\frac{2}{2(1 - \frac{x}{2})} = -\frac{1}{1 - \frac{x}{2}} \quad \text{(valid for } |x| < 2\text{)} \]
\[ \frac{1}{x+1} = \frac{1}{1 - (-x)} \quad \text{(valid for } |x| < 1\text{)} \]

Now we can use the binomial expansion:
\[ -\frac{1}{1 - \frac{x}{2}} = -\sum_{n=0}^{\infty} \left(\frac{x}{2}\right)^n = -\left(1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} + \frac{x^4}{16} + \ldots\right) \]
\[ \frac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^n x^n = 1 - x + x^2 - x^3 + x^4 - \ldots \]

### Step 6: Combining the Expansions
Now we combine the two expansions:
\[ -\left(1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} + \frac{x^4}{16} + \ldots\right) + \left(1 - x + x^2 - x^3 + x^4 - \ldots\right) \]

### Step 7: Finding the Coefficient of \( x^4 \)
The coefficient of \( x^4 \) from the first series is \( -\frac{1}{16} \) and from the second series is \( 1 \). Thus, the total coefficient of \( x^4 \) is:
\[ 1 - \frac{1}{16} = \frac{16}{16} - \frac{1}{16} = \frac{15}{16} \]

### Step 8: Validity of the Expansion
The expansions are valid for:
\[ |x| < 1 \quad \text{and} \quad |x| < 2 \]
Thus, the overall condition is:
\[ |x| < 1 \]

### Final Answer
The coefficient of \( x^4 \) in the expansion of \( \frac{3x}{(x-2)(x+1)} \) is \( \frac{15}{16} \), valid for \( |x| < 1 \).

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