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

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 inside the parentheses
The series \( 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \ldots \) can be recognized as the Taylor series expansion of \( \cosh(x) \):

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

Thus, we can rewrite the expression as:

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

### Step 2: Use the identity for hyperbolic functions
We know that:

\[ \cosh^2(x) = \frac{1 + \cosh(2x)}{2} \]

Applying this identity, we have:

\[ \left(\cosh(\sqrt{x})\right)^2 = \frac{1 + \cosh(2\sqrt{x})}{2} \]

### Step 3: Expand the hyperbolic cosine
Now we need to expand \( \cosh(2\sqrt{x}) \):

\[ \cosh(2\sqrt{x}) = \sum_{k=0}^{\infty} \frac{(2\sqrt{x})^{2k}}{(2k)!} = \sum_{k=0}^{\infty} \frac{4^k x^k}{(2k)!} \]

### Step 4: Substitute back into the expression
Substituting back into our expression gives:

\[ \frac{1 + \sum_{k=0}^{\infty} \frac{4^k x^k}{(2k)!}}{2} = \frac{1}{2} + \frac{1}{2} \sum_{k=0}^{\infty} \frac{4^k x^k}{(2k)!} \]

### Step 5: Identify the coefficient of \( x^n \)
To find the coefficient of \( x^n \) when \( n \) is odd, we note that the series only contains even powers of \( x \) from the \( \cosh \) expansion. Therefore, the coefficient of \( x^n \) in \( \cosh(2\sqrt{x}) \) is zero for all odd \( n \).

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

\[ \boxed{0} \]