Coefficient of `x^(n)` in `(1-2x)/e^x` is

Options

A. 1) `(1+2n)/(n!)`
B. 2) `(-1)^n((1+2n))/(n!)`
C. 3) `n((1-2n))/(n!)`
D. 4) `(-1)^n((1+4n))/(n!)`

Answer

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

### Step 1: Expand \( e^x \)
The exponential function \( e^x \) can be expanded using its Taylor series:
\[ e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!} = 1 + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \]

### Step 2: Write \( \frac{1 - 2x}{e^x} \)
We can rewrite the expression as:
\[ \frac{1 - 2x}{e^x} = (1 - 2x) e^{-x} \]
Now, we need to find the series expansion for \( e^{-x} \).

### Step 3: Expand \( e^{-x} \)
Using the series expansion for \( e^x \), we can find \( e^{-x} \):
\[ e^{-x} = \sum_{k=0}^{\infty} \frac{(-x)^k}{k!} = 1 - \frac{x^1}{1!} + \frac{x^2}{2!} - \frac{x^3}{3!} + \ldots \]

### Step 4: Combine the expansions
Now, we can multiply \( (1 - 2x) \) with the series expansion of \( e^{-x} \):
\[ (1 - 2x) \left( 1 - \frac{x^1}{1!} + \frac{x^2}{2!} - \frac{x^3}{3!} + \ldots \right) \]

### Step 5: Distribute \( (1 - 2x) \)
Distributing \( (1 - 2x) \) gives:
\[ 1 - \frac{x^1}{1!} + \frac{x^2}{2!} - \frac{x^3}{3!} - 2x + 2 \cdot \frac{x^2}{1!} - 2 \cdot \frac{x^3}{2!} + \ldots \]
Now, we can combine like terms:
- The coefficient of \( x^0 \) is \( 1 \).
- The coefficient of \( x^1 \) is \( -1 - 2 = -3 \).
- The coefficient of \( x^2 \) is \( \frac{1}{2!} + 2 \cdot \frac{1}{1!} = \frac{1}{2} + 2 = \frac{5}{2} \).
- The coefficient of \( x^3 \) is \( -\frac{1}{3!} - 2 \cdot \frac{1}{2!} = -\frac{1}{6} - 1 = -\frac{7}{6} \).

### Step 6: Generalize for \( x^n \)
The coefficient of \( x^n \) can be generalized as:
\[ \text{Coefficient of } x^n = -\frac{1}{n!} + 2 \cdot \frac{1}{(n-1)!} \]
This can be simplified to:
\[ -\frac{1}{n!} + \frac{2 \cdot n}{n!} = \frac{-1 + 2n}{n!} \]

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