The coefficient of `x^(n)` in the expansion of `(1-ax-x^(2))/(e^(x))` is

Options

A. `((-1)^(n))/(n!) {-n^(2)-n(a+1)+1}`
B. `((-1)^(n))/(n!) {n^(2)-n(a+1)-1}`
C. `((-1)^(n))/(n!) {-n^(2)+n(a+1)+1}`
D. none

Answer

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

### Step 1: Expand \( e^{-x} \)
We start with the Taylor series expansion of \( e^{-x} \):
\[ e^{-x} = 1 - \frac{x}{1!} + \frac{x^2}{2!} - \frac{x^3}{3!} + \cdots = \sum_{k=0}^{\infty} \frac{(-1)^k x^k}{k!} \]

### Step 2: Rewrite the expression
We can rewrite the expression \( \frac{1 - ax - x^2}{e^x} \) as:
\[ (1 - ax - x^2)e^{-x} \]

### Step 3: Distribute \( e^{-x} \)
Now we will distribute \( e^{-x} \) across the terms in the numerator:
\[ (1)e^{-x} - (ax)e^{-x} - (x^2)e^{-x} \]

### Step 4: Find the coefficient of \( x^n \) from each term
1. **From \( e^{-x} \)**:
The coefficient of \( x^n \) in \( e^{-x} \) is \( \frac{(-1)^n}{n!} \).

2. **From \( -ax e^{-x} \)**:
The coefficient of \( x^n \) in \( -ax e^{-x} \) is given by the coefficient of \( x^{n-1} \) in \( e^{-x} \), multiplied by \( -a \):
\[ -a \cdot \frac{(-1)^{n-1}}{(n-1)!} \]

3. **From \( -x^2 e^{-x} \)**:
The coefficient of \( x^n \) in \( -x^2 e^{-x} \) is given by the coefficient of \( x^{n-2} \) in \( e^{-x} \), multiplied by \( -1 \):
\[ -\frac{(-1)^{n-2}}{(n-2)!} \]

### Step 5: Combine the coefficients
Now we combine the coefficients from all three terms:
\[ \text{Coefficient of } x^n = \frac{(-1)^n}{n!} - a \cdot \frac{(-1)^{n-1}}{(n-1)!} - \frac{(-1)^{n-2}}{(n-2)!} \]

### Step 6: Simplify the expression
Factoring out \( (-1)^{n-2} \):
\[ = (-1)^{n-2} \left( \frac{(-1)^2}{n!} - a \cdot \frac{(-1)}{(n-1)!} - 1 \cdot \frac{1}{(n-2)!} \right) \]
\[ = (-1)^{n-2} \left( \frac{1}{n!} - \frac{a}{(n-1)!} - \frac{1}{(n-2)!} \right) \]

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