If `omega ne 1` is a complex cube root of unity, then `5.23 + omega + omega^(((1)/(2) + (3)/(8) + (9)/(32) + (27)/(128)+...))` is equal to __________

Answer

To solve the expression \( 5.23 + \omega + \omega^{\left(\frac{1}{2} + \frac{3}{8} + \frac{9}{32} + \frac{27}{128} + \ldots\right)} \), where \( \omega \) is a complex cube root of unity, we can follow these steps:

### Step 1: Understanding the properties of \( \omega \)
The complex cube roots of unity satisfy the equations:
1. \( 1 + \omega + \omega^2 = 0 \)
2. \( \omega^3 = 1 \)

From the first equation, we can derive that:
\[ \omega + \omega^2 = -1 \]

### Step 2: Simplifying the exponent
We need to evaluate the series \( \frac{1}{2} + \frac{3}{8} + \frac{9}{32} + \frac{27}{128} + \ldots \).

Notice that the terms can be expressed as:
- The first term is \( \frac{1}{2} = \frac{1^2}{2^1} \)
- The second term is \( \frac{3}{8} = \frac{3^2}{2^3} \)
- The third term is \( \frac{9}{32} = \frac{3^2}{2^5} \)
- The fourth term is \( \frac{27}{128} = \frac{3^3}{2^7} \)

This suggests a pattern where the \( n \)-th term can be expressed as:
\[ \frac{(3/4)^{n-1}}{2^{n-1}} = \frac{3^{n-1}}{2^{n+1}} \]

### Step 3: Identifying the series as a geometric series
The series can be rewritten as:
\[ \sum_{n=1}^{\infty} \frac{(3/4)^{n-1}}{2^{n-1}} = \frac{1/2}{1 - 3/4} = \frac{1/2}{1/4} = 2 \]

### Step 4: Substituting back into the expression
Now we can substitute this result back into our expression:
\[ \omega^{\left(\frac{1}{2} + \frac{3}{8} + \frac{9}{32} + \frac{27}{128} + \ldots\right)} = \omega^{2} \]

### Step 5: Final calculation
Now we can substitute \( \omega^2 \) back into the original expression:
\[ 5.23 + \omega + \omega^2 = 5.23 + \omega - 1 \]
Using the property \( \omega + \omega^2 = -1 \):
\[ = 5.23 - 1 = 4.23 \]

### Final Answer
Thus, the value of the expression is:
\[ \boxed{4.23} \]