Inner Summation Evaluation
The inner summation is $S_{inner} = \sum_{r=1}^{10} \left( \sin \frac{2r\pi}{11} - i \cos \frac{2r\pi}{11} \right)$.
We can rewrite the term inside the summation:
$\sin \frac{2r\pi}{11} - i \cos \frac{2r\pi}{11} = -i \left( \cos \frac{2r\pi}{11} + i \sin \frac{2r\pi}{11} \right) = -i e^{i \frac{2r\pi}{11}}$
Let $\omega = e^{i \frac{2\pi}{11}}$. Then the expression becomes:
$S_{inner} = \sum_{r=1}^{10} -i \omega^r = -i \sum_{r=1}^{10} \omega^r$
Since $\omega$ is an 11th root of unity (i.e., $\omega^{11} = 1$), the sum of all roots is zero: $1 + \omega + \omega^2 + \dots + \omega^{10} = 0$.
Therefore, the sum from $r=1$ to $10$ is $\sum_{r=1}^{10} \omega^r = (\sum_{r=0}^{10} \omega^r) - \omega^0 = 0 - 1 = -1$.
Substituting this back, we get $S_{inner} = -i(-1) = i$.
Outer Summation Calculation
The original expression simplifies to $\sum_{K=1}^{32} (3K+2) (S_{inner})^K = \sum_{K=1}^{32} (3K+2) i^K$.
This is an arithmetic-geometric series. We can use the standard formula for $\sum_{K=1}^{n} (aK+b)x^K$:
$ \sum_{K=1}^{n} (aK+b)x^K = \frac{ax(1-(n+1)x^n+nx^{n+1})}{(1-x)^2} + \frac{bx(1-x^n)}{1-x} $
In this case, $a=3$, $b=2$, $n=32$, and $x=i$. First, calculate the necessary powers of $i$:
- $i^{32} = (i^4)^8 = 1^8 = 1$
- $i^{33} = i^{32} \cdot i = 1 \cdot i = i$
Next, calculate the terms needed for the formula:
- $1-x = 1-i$
- $(1-x)^2 = (1-i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i$
- $1-x^n = 1 - i^{32} = 1 - 1 = 0$
- $1-(n+1)x^n+nx^{n+1} = 1 - (32+1)i^{32} + 32i^{33} = 1 - 33(1) + 32(i) = 1 - 33 + 32i = -32 + 32i = 32(i-1)$
Now, substitute these values into the formula:
$ S_{outer} = \frac{3 \cdot i \cdot (32(i-1))}{(-2i)} + \frac{2 \cdot i \cdot (0)}{(1-i)} $
$ S_{outer} = \frac{96i(i-1)}{-2i} + 0 $
Simplify the expression:
$ S_{outer} = -48(i-1) $
$ S_{outer} = -48i + 48 = 48(1-i) $
Final Expression Result
The value of the given expression is $48(1-i)$.
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