The sum of the series `1+(3^(2))/(2!)+(3^(4))/(4!)+(3^(6))/(6!)+…` to `oo` is

Options

A. `e^(-3)`
B. `e^(3)`
C. `(1)/(2)(e^(3)-e^(-3))`
D. `(1)/(2)(e^(3)+e^(-3))`

Answer

To find the sum of the series

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

we can recognize that this series resembles the Taylor series expansion of the exponential function.

### Step-by-Step Solution:

1. **Identify the Pattern**:
The series can be rewritten as:
\[ S = \sum_{n=0}^{\infty} \frac{3^{2n}}{(2n)!} \]
This is because the terms of the series are of the form \( \frac{3^{2n}}{(2n)!} \).

2. **Relate to the Exponential Function**:
The Taylor series expansion for \( e^x \) is given by:
\[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \]
To relate our series to this, we can use the fact that the series for \( e^x \) includes both even and odd powers of \( x \).

3. **Use the Even Function**:
The series for \( e^x \) can be split into even and odd parts:
\[ e^x = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} + \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!} \]
The first part corresponds to our series when \( x = 3 \):
\[ \sum_{n=0}^{\infty} \frac{3^{2n}}{(2n)!} = e^3 + e^{-3} \]

4. **Calculate the Sum**:
The sum of the series we have is:
\[ S = \frac{e^3 + e^{-3}}{2} \]
This is because the series we derived corresponds to the hyperbolic cosine function:
\[ \cosh(x) = \frac{e^x + e^{-x}}{2} \]
Thus, we can write:
\[ S = \cosh(3) \]

### Final Result:
The sum of the series is:
\[ S = \cosh(3) \]