The coefficient 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 1: Rewrite the expression
We start with the expression:
\[ \frac{1 - 2x - x^2}{e^{-x}} = (1 - 2x - x^2) \cdot e^x \]

### Step 2: Expand \( e^x \)
The exponential function \( e^x \) can be expanded using its Taylor series:
\[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \]

### Step 3: Distribute \( (1 - 2x - x^2) \)
Now we distribute \( (1 - 2x - x^2) \) over the series expansion of \( e^x \):
\[ (1 - 2x - x^2) \cdot e^x = e^x - 2x e^x - x^2 e^x \]

### Step 4: Find the coefficient of \( x^k \)
We need to find the coefficient of \( x^k \) in the expression:
\[ e^x - 2x e^x - x^2 e^x \]

1. **From \( e^x \)**: The coefficient of \( x^k \) is \( \frac{1}{k!} \).
2. **From \( -2x e^x \)**: The coefficient of \( x^k \) is \( -2 \cdot \frac{1}{(k-1)!} \) (since we need \( x^{k-1} \)).
3. **From \( -x^2 e^x \)**: The coefficient of \( x^k \) is \( -\frac{1}{(k-2)!} \) (since we need \( x^{k-2} \)).

### Step 5: Combine the coefficients
Now, we combine the coefficients from each term:
\[ \text{Coefficient of } x^k = \frac{1}{k!} - 2 \cdot \frac{1}{(k-1)!} - \frac{1}{(k-2)!} \]

### Step 6: Simplify the expression
To combine these fractions, we can find a common denominator, which is \( k! \):
\[ \text{Coefficient of } x^k = \frac{1}{k!} - \frac{2k}{k!} - \frac{k(k-1)}{k!} \]
\[ = \frac{1 - 2k - k(k-1)}{k!} \]
\[ = \frac{1 - 2k - k^2 + k}{k!} \]
\[ = \frac{1 - k - k^2}{k!} \]

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