If n is odd, then Coefficient of `x^n` in `(1+ (x^2)/(2!) +x^4/(4!) + ....oo)^2 = `

Options

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

Answer

To find the coefficient of \( x^n \) in the expression

\[ \left(1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \ldots \right)^2 \]

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 \( \cosh(x) \), which is given by:

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

Thus, we can rewrite our expression as:

\[ \left(\cosh(x)\right)^2 \]

### Step 2: Use the identity for hyperbolic functions
Using the identity for hyperbolic functions, we have:

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

So we can express our original series as:

\[ \left(1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \ldots \right)^2 = \frac{1 + \cosh(2x)}{2} \]

### Step 3: Expand \( \cosh(2x) \)
The series for \( \cosh(2x) \) is:

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

### Step 4: Identify the powers of \( x \)
From the expansion of \( \cosh(2x) \), we can see that it contains only even powers of \( x \). Therefore, when we take the expression \( \frac{1 + \cosh(2x)}{2} \), it will also contain only even powers of \( x \).

### Step 5: Conclude about odd coefficients
Since the expansion of \( \frac{1 + \cosh(2x)}{2} \) contains only even powers of \( x \), it follows that there are no terms involving odd powers of \( x \). Thus, the coefficient of \( x^n \) for any odd \( n \) is zero.

### Final Answer
Therefore, the coefficient of \( x^n \) in the given expression when \( n \) is odd is:

\[ \text{Coefficient of } x^n = 0 \]