`(1+(2^(2))/(2!) + (2^(4))/(3!) +(2^(6))/(4!)+...)/(1+(1)/(2!)+(2)/(3!)+(2^(2))/(4!)+...)`

Options

A. `e^(2) + 1`
B. `e^(2)-1`
C. `(e-1)//(e+1)`
D. None

Answer

To solve the given expression

\[ L = \frac{1 + \frac{2^2}{2!} + \frac{2^4}{3!} + \frac{2^6}{4!} + \ldots}{1 + \frac{1}{2!} + \frac{2}{3!} + \frac{2^2}{4!} + \ldots} \]

we will analyze both the numerator and the denominator separately.

### Step 1: Analyze the Numerator

The numerator can be expressed as:

\[ N = 1 + \frac{2^2}{2!} + \frac{2^4}{3!} + \frac{2^6}{4!} + \ldots \]

This series can be recognized as a Taylor series expansion of \( e^{x} \) where \( x = 2^2 = 4 \).

To express it in terms of \( e^x \):

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

This can be rewritten by changing the index of summation:

\[ N = \sum_{m=1}^{\infty} \frac{4^{m-1}}{m!} = \frac{1}{4} \sum_{m=1}^{\infty} \frac{4^m}{m!} = \frac{1}{4} (e^4 - 1) \]

### Step 2: Analyze the Denominator

The denominator can be expressed as:

\[ D = 1 + \frac{1}{2!} + \frac{2}{3!} + \frac{2^2}{4!} + \ldots \]

This series can be recognized as:

\[ D = \sum_{n=0}^{\infty} \frac{2^n}{(n+1)!} \]

Similar to the numerator, we can change the index of summation:

\[ D = \sum_{m=1}^{\infty} \frac{2^{m-1}}{m!} = \frac{1}{2} \sum_{m=1}^{\infty} \frac{2^m}{m!} = \frac{1}{2} (e^2 - 1) \]

### Step 3: Combine the Results

Now we can substitute \( N \) and \( D \) back into our expression for \( L \):

\[ L = \frac{N}{D} = \frac{\frac{1}{4}(e^4 - 1)}{\frac{1}{2}(e^2 - 1)} = \frac{1}{4} \cdot \frac{2(e^4 - 1)}{(e^2 - 1)} = \frac{1}{2} \cdot \frac{e^4 - 1}{e^2 - 1} \]

### Step 4: Simplify Further

Using the identity \( a^2 - b^2 = (a - b)(a + b) \):

\[ e^4 - 1 = (e^2 - 1)(e^2 + 1) \]

Thus, we can simplify:

\[ L = \frac{1}{2} \cdot \frac{(e^2 - 1)(e^2 + 1)}{(e^2 - 1)} = \frac{1}{2}(e^2 + 1) \]

### Final Result

The final result is:

\[ L = \frac{1}{2}(e^2 + 1) \]