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

Answer

To find the coefficient of \( x^n \) in the expansion of \( \frac{x}{(2x+1)(x-2)} \), we will use the method of partial fractions and then find the required coefficient.

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

### Step 2: Clear the denominators
Multiply both sides by \( (2x+1)(x-2) \):
\[ x = A(x-2) + B(2x+1) \]

### Step 3: Expand and collect like terms
Expanding the right side gives:
\[ x = Ax - 2A + 2Bx + B \]
Combining like terms:
\[ x = (A + 2B)x + (-2A + B) \]

### Step 4: Set up equations for coefficients
Now, we equate coefficients from both sides:
1. From the coefficient of \( x \):
\[ A + 2B = 1 \]
2. From the constant term:
\[ -2A + B = 0 \]

### Step 5: Solve the system of equations
From the second equation, we can express \( B \) in terms of \( A \):
\[ B = 2A \]
Substituting \( B \) into the first equation:
\[ A + 2(2A) = 1 \implies A + 4A = 1 \implies 5A = 1 \implies A = \frac{1}{5} \]
Now substituting \( A \) back to find \( B \):
\[ B = 2A = 2 \cdot \frac{1}{5} = \frac{2}{5} \]

### Step 6: Write the partial fraction decomposition
Now we have:
\[ \frac{x}{(2x+1)(x-2)} = \frac{1/5}{2x+1} + \frac{2/5}{x-2} \]

### Step 7: Expand each term using series
We can expand each term using the geometric series:
1. For \( \frac{1/5}{2x+1} \):
\[ \frac{1/5}{2x+1} = \frac{1/5}{1 - (-2x)} = \frac{1}{5} \sum_{n=0}^{\infty} (-2x)^n = \frac{1}{5} \sum_{n=0}^{\infty} (-2)^n x^n \]
This series converges for \( |2x| < 1 \) or \( |x| < \frac{1}{2} \).

2. For \( \frac{2/5}{x-2} \):
\[ \frac{2/5}{x-2} = \frac{2/5}{-2(1 - \frac{x}{2})} = -\frac{1}{5} \sum_{n=0}^{\infty} \left(\frac{x}{2}\right)^n \]
This series converges for \( |x| < 2 \).

### Step 8: Combine the series
Combining both series, we have:
\[ \frac{1}{5} \sum_{n=0}^{\infty} (-2)^n x^n - \frac{1}{5} \sum_{n=0}^{\infty} \left(\frac{x}{2}\right)^n \]

### Step 9: Find the coefficient of \( x^n \)
The coefficient of \( x^n \) from the first series is \( \frac{(-2)^n}{5} \) and from the second series is \( -\frac{1}{5} \cdot \frac{1}{2^n} \). Thus, the total coefficient of \( x^n \) is:
\[ \frac{1}{5} \left( (-2)^n - \frac{1}{2^n} \right) = \frac{1}{5} \left( (-2)^n - \frac{1}{2^n} \right) \]

### Step 10: Specify the interval of validity
The expansion is valid for \( |x| < \frac{1}{2} \) (from the first series) and \( |x| < 2 \) (from the second series). Therefore, the overall interval of validity is:
\[ |x| < \frac{1}{2} \]

### Final Answer
The coefficient of \( x^n \) in the expansion is:
\[ \frac{1}{5} \left( (-2)^n - \frac{1}{2^n} \right) \]
and the interval of validity is \( |x| < \frac{1}{2} \).