The coefficient of `x^(n)` in the expansion of `(1 + (x^(2))/(2!) + (x^(4))/(4!) + ...)^(2)` , when n is odd is

Options

A. `(2^(2n))/((2n)!)`
B. `(2^(n))/(n!)`
C. 0
D. none

Answer

To find the coefficient of \( x^n \) in the expansion of \( \left( 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \ldots \right)^2 \) when \( n \) is odd, we can follow these steps:

### Step 1: Recognize the series
The series \( 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \ldots \) can be recognized as the Taylor series expansion for \( e^x \) evaluated at \( x^2 \). Thus, we can rewrite it as:
\[ 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \ldots = e^{x^2} \]

### Step 2: Rewrite the expression
Now, we can rewrite the original expression:
\[ \left( 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \ldots \right)^2 = \left( e^{x^2} \right)^2 = e^{2x^2} \]

### Step 3: Analyze the expansion
The function \( e^{2x^2} \) can be expanded as:
\[ e^{2x^2} = \sum_{k=0}^{\infty} \frac{(2x^2)^k}{k!} = \sum_{k=0}^{\infty} \frac{2^k x^{2k}}{k!} \]
This series contains only even powers of \( x \) because \( x^{2k} \) is always even.

### Step 4: Identify the coefficient of \( x^n \)
Since we are interested in the coefficient of \( x^n \) where \( n \) is odd, we note that there are no odd powers of \( x \) in the expansion of \( e^{2x^2} \).

### Conclusion
Thus, the coefficient of \( x^n \) in the expansion when \( n \) is odd is:
\[ \text{Coefficient} = 0 \]