`1 + ((log n)^(2))/(2!) + ((log n)^(4))/(4!) + ...oo=`

Options

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

Answer

To solve the given series \( S = 1 + \frac{(\log n)^2}{2!} + \frac{(\log n)^4}{4!} + \ldots \), we can recognize that this series resembles the expansion of the hyperbolic cosine function.

### Step 1: Recognize the series
The series can be rewritten as:
\[ S = \sum_{k=0}^{\infty} \frac{(\log n)^{2k}}{(2k)!} \]
This is the Taylor series expansion for \( \cosh(x) \), where \( x = \log n \).

### Step 2: Use the hyperbolic cosine function
The hyperbolic cosine function is defined as:
\[ \cosh(x) = \frac{e^x + e^{-x}}{2} \]
Thus, substituting \( x = \log n \):
\[ S = \cosh(\log n) \]

### Step 3: Simplify using properties of logarithms and exponentials
Using the property of hyperbolic cosine:
\[ \cosh(\log n) = \frac{e^{\log n} + e^{-\log n}}{2} \]
Since \( e^{\log n} = n \) and \( e^{-\log n} = \frac{1}{n} \), we can substitute these values:
\[ S = \frac{n + \frac{1}{n}}{2} \]

### Step 4: Final expression
Thus, we can write:
\[ S = \frac{n + \frac{1}{n}}{2} \]

### Conclusion
The final value of the given series is:
\[ \boxed{\frac{n + \frac{1}{n}}{2}} \]