`1+(omega^2x)/(1!)+(omegax^2)/(2!)+(x^3)/(3!) + (omega^2x^4)/(4!) +(omegax^5)/(5!) +(x^6)/(6!) + .... oo = `

Options

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

Answer

To solve the series

\[ S = 1 + \frac{\omega^2 x}{1!} + \frac{\omega x^2}{2!} + \frac{x^3}{3!} + \frac{\omega^2 x^4}{4!} + \frac{\omega x^5}{5!} + \frac{x^6}{6!} + \ldots \]

where \(\omega\) is a cube root of unity, we can follow these steps:

### Step 1: Understanding the Series
The series can be rewritten by observing the pattern in the coefficients. The terms alternate between \(\omega^2\), \(\omega\), and \(1\) based on the power of \(x\).

### Step 2: Recognizing the Pattern
The coefficients of the series can be grouped based on the powers of \(x\):
- For \(n = 0\): \(1\)
- For \(n = 1\): \(\frac{\omega^2 x}{1!}\)
- For \(n = 2\): \(\frac{\omega x^2}{2!}\)
- For \(n = 3\): \(\frac{x^3}{3!}\)
- For \(n = 4\): \(\frac{\omega^2 x^4}{4!}\)
- For \(n = 5\): \(\frac{\omega x^5}{5!}\)
- For \(n = 6\): \(\frac{x^6}{6!}\)
- And so on...

### Step 3: Using the Exponential Series
We know that the exponential function can be expressed as:

\[ e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \]

We can use this to express the series in terms of \(e^{\omega^2 x}\) and \(e^{\omega x}\).

### Step 4: Grouping Terms
We can express the series as:

\[ S = e^{\omega^2 x} + e^{\omega x} + e^{x} \]

However, we need to recognize that the series can be simplified by considering the periodicity of \(\omega\).

### Step 5: Final Expression
Since \(\omega^3 = 1\), we can conclude that the series converges to:

\[ S = e^{\omega^2 x} \]

### Conclusion
Thus, the value of the series is:

\[ S = e^{\omega^2 x} \]