The coefficent of `x^(n)` in the expansion of `(1+(x^(2))/(2!)+(x^(4))/(4!)+…)^(2)`
When n is odd is

Options

A. `(2^(n))/(n!)`
B. `(2^(n))/(2n!)`
C. 0
D. `(2^(2x))/(n!)`

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 will follow these steps:

### Step 1: Recognize the series
The series \( 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \ldots \) is the Taylor series expansion for \( \cosh(x) \).

\[ \cosh(x) = \sum_{k=0}^{\infty} \frac{x^{2k}}{(2k)!} \]

Thus, we can rewrite our expression as:

\[ \left( \cosh(\sqrt{x}) \right)^2 \]

### Step 2: Use the identity of hyperbolic functions
Using the identity for \( \cosh(x) \):

\[ \cosh(x) = \frac{e^x + e^{-x}}{2} \]

We have:

\[ \cosh(\sqrt{x}) = \frac{e^{\sqrt{x}} + e^{-\sqrt{x}}}{2} \]

Squaring this gives:

\[ \left( \cosh(\sqrt{x}) \right)^2 = \frac{(e^{\sqrt{x}} + e^{-\sqrt{x}})^2}{4} = \frac{e^{2\sqrt{x}} + 2 + e^{-2\sqrt{x}}}{4} \]

### Step 3: Expand the expression
Now we need to expand \( e^{2\sqrt{x}} \) and \( e^{-2\sqrt{x}} \):

\[ e^{2\sqrt{x}} = \sum_{k=0}^{\infty} \frac{(2\sqrt{x})^k}{k!} = \sum_{k=0}^{\infty} \frac{2^k x^{k/2}}{k!} \]

\[ e^{-2\sqrt{x}} = \sum_{k=0}^{\infty} \frac{(-2\sqrt{x})^k}{k!} = \sum_{k=0}^{\infty} \frac{(-2)^k x^{k/2}}{k!} \]

### Step 4: Combine the expansions
Combining these, we get:

\[ e^{2\sqrt{x}} + e^{-2\sqrt{x}} = \sum_{k=0}^{\infty} \left( \frac{2^k + (-2)^k}{k!} \right) x^{k/2} \]

### Step 5: Find the coefficient of \( x^n \)
Now, we need to find the coefficient of \( x^n \) in the expression:

\[ \frac{1}{4} \left( e^{2\sqrt{x}} + e^{-2\sqrt{x}} + 2 \right) \]

Since \( n \) is odd, the terms \( 2^k + (-2)^k \) will cancel out for odd \( k \), leading to:

\[ \frac{1}{4} \cdot 0 = 0 \]

### Conclusion
Thus, the coefficient of \( x^n \) in the expansion when \( n \) is odd is:

\[ \boxed{0} \]