Step 1: Understanding the Concept:
This problem involves two nested summations. The inner sum involves complex roots of unity, while the outer sum is an Arithmetic-Geometric Progression (AGP). We need to solve the inner sum first and then use the resulting common ratio to sum the outer series.
Step 2: Key Formula or Approach:
1. Let \( \omega = e^{i 2\pi/11} \) be an 11th root of unity. The sum of all roots is \( \sum_{r=0}^{n-1} \omega^r = 0 \).
2. The outer sum is of the form \( \sum (ak + b)r^k \).
Step 3: Detailed Explanation:
Let's evaluate the inner sum \( S_{in} = \sum_{r=1}^{10} \left( \sin \frac{2r\pi}{11} - i \cos \frac{2r\pi}{11} \right) \).
Factor out \( -i \) from the terms:
\[ S_{in} = -i \sum_{r=1}^{10} \left( \cos \frac{2r\pi}{11} + i \sin \frac{2r\pi}{11} \right) \]
Using Euler's formula \( e^{i\theta} = \cos \theta + i \sin \theta \):
\[ S_{in} = -i \sum_{r=1}^{10} e^{i \frac{2r\pi}{11}} \]
Let \( \omega = e^{i \frac{2\pi}{11}} \). Then \( S_{in} = -i (\omega^1 + \omega^2 + \dots + \omega^{10}) \).
We know that the sum of all 11th roots of unity is \( 1 + \omega + \omega^2 + \dots + \omega^{10} = 0 \).
Therefore, \( \omega + \omega^2 + \dots + \omega^{10} = -1 \).
Substituting this back:
\[ S_{in} = -i (-1) = i \]
Now substitute this into the outer sum:
\[ S = \sum_{K=1}^{32} (3K + 2) i^K \]
This is an AGP. Let's write out some terms:
\[ S = 5i + 8i^2 + 11i^3 + 14i^4 + 17i^5 + \dots + 98i^{32} \]
Group the terms in sets of 4 (since \( i^1, i^2, i^3, i^4 \) repeat in pattern):
Set 1: \( 5i - 8 - 11i + 14 = 6 - 6i \).
Set 2: \( 17i - 20 - 23i + 26 = 6 - 6i \).
There are 32 terms in total, which means there are \( 32/4 = 8 \) sets of such four terms.
Each set yields the same value \( 6(1 - i) \).
Total Sum \( S = 8 \times 6(1 - i) = 48(1 - i) \).
Step 4: Final Answer:
The inner summation simplifies to \( i \) and the outer summation of the AGP results in \( 48(1 - i) \). This matches option (B).