`4/(1!) +16/(3!)+(64)/(5!) + ......= `

Options

A. `e +1/e`
B. `e -1/e`
C. `e^2 + 1/e^2`
D. `e^(2) -1/e^2`

Answer

To solve the expression \( S = \frac{4}{1!} + \frac{16}{3!} + \frac{64}{5!} + \ldots \), we can identify a pattern in the terms.

### Step 1: Identify the general term
The terms can be expressed as:
- The first term is \( \frac{4}{1!} = \frac{2^2}{1!} \)
- The second term is \( \frac{16}{3!} = \frac{4^2}{3!} = \frac{(2^2)^2}{3!} \)
- The third term is \( \frac{64}{5!} = \frac{8^2}{5!} = \frac{(2^3)^2}{5!} \)

We can see that the \( n \)-th term (for odd \( n \)) can be expressed as:
\[ \frac{(2^{n})^2}{(n)!} \]
where \( n \) takes odd values (1, 3, 5, ...).

### Step 2: Write the series in summation form
The series can be rewritten as:
\[ S = \sum_{k=0}^{\infty} \frac{(2^{2k})}{(2k+1)!} \]
This represents the sum of the series for odd factorial denominators.

### Step 3: Relate to exponential functions
Recall the Taylor series expansion for \( e^x \):
\[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \]
We can use this to find the series for \( e^{2} \) and \( e^{-2} \).

### Step 4: Calculate \( e^2 \) and \( e^{-2} \)
The series for \( e^2 \) is:
\[ e^2 = \sum_{n=0}^{\infty} \frac{2^n}{n!} \]
The series for \( e^{-2} \) is:
\[ e^{-2} = \sum_{n=0}^{\infty} \frac{(-2)^n}{n!} \]

### Step 5: Combine the series
By combining \( e^2 \) and \( e^{-2} \):
\[ e^2 + e^{-2} = 2 + \sum_{k=1}^{\infty} \left( \frac{2^{2k}}{(2k)!} + \frac{(-2)^{2k}}{(2k)!} \right) \]
The odd terms will cancel out, leaving us with:
\[ e^2 - e^{-2} = 2 \sum_{k=0}^{\infty} \frac{(2^{2k})}{(2k+1)!} \]

### Step 6: Solve for \( S \)
Thus, we have:
\[ S = \frac{1}{2} (e^2 - e^{-2}) \]

### Final Result
Using the identity \( e^x - e^{-x} = 2\sinh(x) \), we find:
\[ S = \frac{1}{2} (e^2 - e^{-2}) = \sinh(2) \]

### Conclusion
The value of the series \( S = \frac{4}{1!} + \frac{16}{3!} + \frac{64}{5!} + \ldots \) is:
\[ S = \sinh(2) \]