For `|x| lt 1/5`, the coefficient of `x^(3)` in the expansion of `(1)/((1-5x)(1-4x))` is

Options

A. 369
B. 370
C. 371
D. 372

Answer

To find the coefficient of \( x^3 \) in the expansion of \( \frac{1}{(1-5x)(1-4x)} \) for \( |x| < \frac{1}{5} \), we will use partial fractions and the binomial series expansion.

### Step-by-step Solution:

1. **Set Up the Partial Fraction Decomposition:**
We start with the expression:
\[ \frac{1}{(1-5x)(1-4x)} = \frac{A}{1-5x} + \frac{B}{1-4x} \]
where \( A \) and \( B \) are constants we need to determine.

2. **Combine the Right Side:**
To combine the right side, we find a common denominator:
\[ \frac{A(1-4x) + B(1-5x)}{(1-5x)(1-4x)} = \frac{1}{(1-5x)(1-4x)} \]
This gives us the equation:
\[ 1 = A(1-4x) + B(1-5x) \]

3. **Expand and Collect Like Terms:**
Expanding the right side:
\[ 1 = A - 4Ax + B - 5Bx \]
Combining like terms, we have:
\[ 1 = (A + B) + (-4A - 5B)x \]

4. **Set Up the System of Equations:**
From the equation \( 1 = (A + B) + (-4A - 5B)x \), we can equate coefficients:
- For the constant term: \( A + B = 1 \)
- For the coefficient of \( x \): \( -4A - 5B = 0 \)

5. **Solve the System of Equations:**
From \( -4A - 5B = 0 \), we can express \( B \) in terms of \( A \):
\[ B = -\frac{4}{5}A \]
Substituting this into \( A + B = 1 \):
\[ A - \frac{4}{5}A = 1 \implies \frac{1}{5}A = 1 \implies A = 5 \]
Then substituting back to find \( B \):
\[ B = 1 - A = 1 - 5 = -4 \]

6. **Rewrite the Original Expression:**
Now we can rewrite the original expression:
\[ \frac{1}{(1-5x)(1-4x)} = \frac{5}{1-5x} - \frac{4}{1-4x} \]

7. **Apply the Binomial Series Expansion:**
The binomial series expansion for \( \frac{1}{1 - kx} \) is:
\[ \frac{1}{1 - kx} = \sum_{n=0}^{\infty} k^n x^n \]
Thus, we have:
\[ \frac{5}{1-5x} = 5 \sum_{n=0}^{\infty} (5x)^n = 5(1 + 5x + 25x^2 + 125x^3 + \ldots) \]
and
\[ -\frac{4}{1-4x} = -4 \sum_{n=0}^{\infty} (4x)^n = -4(1 + 4x + 16x^2 + 64x^3 + \ldots) \]

8. **Find the Coefficient of \( x^3 \):**
Now we find the coefficient of \( x^3 \):
- From \( \frac{5}{1-5x} \): The coefficient of \( x^3 \) is \( 5 \cdot 125 = 625 \).
- From \( -\frac{4}{1-4x} \): The coefficient of \( x^3 \) is \( -4 \cdot 64 = -256 \).

Therefore, the total coefficient of \( x^3 \) is:
\[ 625 - 256 = 369 \]

### Final Answer:
The coefficient of \( x^3 \) in the expansion is \( \boxed{369} \).