The expansion of `[1 + (x^(2))/(2!) + (x^(4))/(4!) + ...]^(2)` in ascending powers of x is

Options

A. `1 + (x^(2))/(2!) + (x^(4))/(4!) + (x^(6))/(6!) +...`
B. `1 + (2^(2)x^(2))/(2!) + (2^(4)x^(4))/(4!) +...`
C. `1 + (2x^(2))/(2!) + (2^(3)x^(4))/(4!) + (2^(5)x^(6))/(6!) + ...`
D. None of these

Answer

To solve the problem, we need to find the expansion of the expression \( \left[ 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \ldots \right]^2 \) in ascending powers of \( x \).

### Step 1: Recognize the Series
The series inside the brackets can be recognized as the Taylor series expansion for the hyperbolic cosine function, \( \cosh(x) \):
\[ \cosh(x) = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \ldots \]
Thus, we can rewrite our expression as:
\[ \left[ \cosh(x) \right]^2 \]

### Step 2: Use the Identity for Hyperbolic Cosine
Using the identity for the square of the hyperbolic cosine:
\[ \cosh^2(x) = \frac{1 + \cosh(2x)}{2} \]
we can substitute this into our expression:
\[ \left[ \cosh(x) \right]^2 = \frac{1 + \cosh(2x)}{2} \]

### Step 3: Expand \( \cosh(2x) \)
Now, we need to expand \( \cosh(2x) \):
\[ \cosh(2x) = 1 + \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} + \ldots = 1 + \frac{4x^2}{2} + \frac{16x^4}{24} + \ldots \]
This simplifies to:
\[ \cosh(2x) = 1 + 2x^2 + \frac{2}{3}x^4 + \ldots \]

### Step 4: Substitute Back into the Expression
Now, substitute \( \cosh(2x) \) back into our expression:
\[ \frac{1 + \cosh(2x)}{2} = \frac{1 + \left( 1 + 2x^2 + \frac{2}{3}x^4 + \ldots \right)}{2} \]
This simplifies to:
\[ \frac{2 + 2x^2 + \frac{2}{3}x^4 + \ldots}{2} = 1 + x^2 + \frac{1}{3}x^4 + \ldots \]

### Step 5: Write the Final Expansion
Thus, the expansion of \( \left[ 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \ldots \right]^2 \) in ascending powers of \( x \) is:
\[ 1 + x^2 + \frac{1}{3}x^4 + \ldots \]

### Final Answer
The final answer in ascending powers of \( x \) is:
\[ 1 + x^2 + \frac{1}{3}x^4 + \ldots \]