The coeffiecinet of `x^k` in the expansion of `(1-2x -x^2)/(e^(-x))` is

Options

A. `(1-k-k^2)/(k!)`
B. `(k^2+1)/(k!)`
C. `(1-k)/(k!)`
D. `1/(k!)`

Answer

To find the coefficient of \( x^k \) in the expansion of \( \frac{1 - 2x - x^2}{e^{-x}} \), we can follow these steps:

### Step-by-Step Solution:

1. **Rewrite the Expression**:
We start with the expression:
\[ \frac{1 - 2x - x^2}{e^{-x}} = (1 - 2x - x^2) e^{x} \]

2. **Expand \( e^x \)**:
The Taylor series expansion of \( e^x \) is:
\[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \]

3. **Multiply the Series**:
Now, we need to multiply \( (1 - 2x - x^2) \) with the series expansion of \( e^x \):
\[ (1 - 2x - x^2) \left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots\right) \]

4. **Distributing the Terms**:
We distribute \( (1 - 2x - x^2) \) across the series:
- The term \( 1 \) contributes:
\[ 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \]
- The term \( -2x \) contributes:
\[ -2x - 2x^2 - \frac{2x^3}{2!} - \frac{2x^4}{3!} - \ldots \]
- The term \( -x^2 \) contributes:
\[ -x^2 - \frac{x^3}{2!} - \frac{x^4}{3!} - \ldots \]

5. **Combine Like Terms**:
Now we combine the contributions for \( x^k \):
- For \( k = 0 \): Coefficient is \( 1 \).
- For \( k = 1 \): Coefficient is \( 1 - 2 = -1 \).
- For \( k = 2 \): Coefficient is \( \frac{1}{2!} - 2 - 1 = \frac{1 - 4 - 2}{2} = \frac{-5}{2} \).
- For \( k \geq 3 \): The general term can be derived as:
\[ \text{Coefficient of } x^k = \frac{1}{k!} - 2 \cdot \frac{1}{(k-1)!} - \frac{1}{(k-2)!} \]

6. **Final Expression**:
We can express the coefficient of \( x^k \) as:
\[ \text{Coefficient of } x^k = \frac{1 - 2(k-1) - (k-2)}{k!} = \frac{1 - 2k + 2 - k + 2}{k!} = \frac{1 - k - k^2}{k!} \]

### Conclusion:
The coefficient of \( x^k \) in the expansion of \( \frac{1 - 2x - x^2}{e^{-x}} \) is:
\[ \frac{1 - k - k^2}{k!} \]