If `(e^(x) -1)^(2) =a_0 +a_(1)x +a_(2) x^2 + ......oo` then `a_4` =

Options

A. `5/12`
B. `7/12`
C. `9/7`
D. 0

Answer

To find the coefficient \( a_4 \) in the expansion of \( (e^x - 1)^2 \), we will follow these steps:

### Step 1: Expand \( (e^x - 1)^2 \)
Using the formula for squaring a binomial, we have:
\[ (e^x - 1)^2 = e^{2x} - 2e^x + 1 \]

### Step 2: Expand \( e^{2x} \) and \( e^x \)
The series expansion for \( e^x \) is:
\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \ldots \]
Thus, for \( e^{2x} \):
\[ e^{2x} = 1 + 2x + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \frac{(2x)^4}{4!} + \ldots = 1 + 2x + \frac{4x^2}{2} + \frac{8x^3}{6} + \frac{16x^4}{24} + \ldots \]
This simplifies to:
\[ e^{2x} = 1 + 2x + 2x^2 + \frac{4}{3}x^3 + \frac{2}{3}x^4 + \ldots \]

For \( e^x \):
\[ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \ldots \]

### Step 3: Substitute into the expansion
Now substituting back into our expression:
\[ (e^x - 1)^2 = (1 + 2x + 2x^2 + \frac{4}{3}x^3 + \frac{2}{3}x^4 + \ldots) - 2(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \ldots) + 1 \]

### Step 4: Combine like terms
Now, we will combine the terms:
- The constant terms: \( 1 - 2 + 1 = 0 \)
- The linear terms: \( 2x - 2x = 0 \)
- The quadratic terms: \( 2x^2 - 2 \cdot \frac{x^2}{2} = 2x^2 - x^2 = x^2 \)
- The cubic terms: \( \frac{4}{3}x^3 - 2 \cdot \frac{x^3}{6} = \frac{4}{3}x^3 - \frac{1}{3}x^3 = \frac{3}{3}x^3 = x^3 \)
- The quartic terms: \( \frac{2}{3}x^4 - 2 \cdot \frac{x^4}{24} = \frac{2}{3}x^4 - \frac{1}{12}x^4 \)

### Step 5: Simplify the quartic term
To simplify the quartic term:
\[ \frac{2}{3} - \frac{1}{12} = \frac{8}{12} - \frac{1}{12} = \frac{7}{12} \]
Thus, the quartic term is:
\[ \frac{7}{12}x^4 \]

### Step 6: Identify \( a_4 \)
From the expansion, we see that:
\[ a_4 = \frac{7}{12} \]

### Final Answer
Thus, the coefficient \( a_4 \) is:
\[ \boxed{\frac{7}{12}} \]