`1+(omegax)/(1!) +(omega^2x^2)/(2!)+(omega^3x^3)/(3!)+(omega^4x^4)/(4!)+(omega^5x^5)/(5!)+ .....oo =`

Options

A. `e^x`
B. `e^(omegax)`
C. `e^(omega^(2)x)`
D. `e^(x^2)`

Answer

To solve the given series:

\[ S = 1 + \frac{\omega x}{1!} + \frac{\omega^2 x^2}{2!} + \frac{\omega^3 x^3}{3!} + \frac{\omega^4 x^4}{4!} + \frac{\omega^5 x^5}{5!} + \ldots \]

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

### Step-by-Step Solution:

1. **Identify the Series**:
The series can be rewritten as:
\[ S = \sum_{n=0}^{\infty} \frac{(\omega x)^n}{n!} \]
This is the general form of the Taylor series for \( e^y \) where \( y = \omega x \).

2. **Recall the Exponential Function**:
The Taylor series expansion for \( e^y \) is given by:
\[ e^y = \sum_{n=0}^{\infty} \frac{y^n}{n!} \]
By substituting \( y = \omega x \), we get:
\[ e^{\omega x} = \sum_{n=0}^{\infty} \frac{(\omega x)^n}{n!} \]

3. **Conclusion**:
Therefore, we can conclude that:
\[ S = e^{\omega x} \]

Thus, the value of the given series is:
\[ \boxed{e^{\omega x}} \]