Question

The expression \[ \sum_{K=1}^{32}(3K+2)\left\{\sum_{r=1}^{10}\left(\sin\frac{2r\pi}{11}-i\cos\frac{2r\pi}{11}\right)\right\}^{K} \] represents:

• A. $48 (1+i)$
• B. $48 (1-i)$
• C. $-\frac{48}{11}(1-i)$
• D. $48(1-i)$

Hint:

Whenever you encounter an index tracking loop over consecutive powers of $i$, always count the total number of terms. If the total number of terms is a multiple of 4, the sum is automatically 0. This saves you from expanding long polynomial terms midway through your calculations.

Answer 1

The Correct Option is B

Concept: This hybrid expression contains an inner trigonometric sequence involving the roots of unity, embedded within an outer Arithmetico-Geometric Progression (AGP). We use Euler's identity ($e^{i\theta} = \cos\theta + i\sin\theta$) to simplify the complex roots of unity, and standard series methods to solve the AGP. Step 1: Simplify the inner complex summation term.
Let the inner term be defined as $S = \sum_{r=1}^{10}\left(\sin\frac{2r\pi}{11}-i \cos\frac{2r\pi}{11}\right)$. Factor out $-i$ to match Euler's form: $$\sin\theta - i\cos\theta = -i(\cos\theta + i\sin\theta) = -i e^{i\theta}$$ $$S = -i \sum_{r=1}^{10} e^{i\frac{2r\pi}{11}}$$ The summation represents the sum of the first 10 complex 11th roots of unity. Recall the identity for the complete sum of all $N$-th roots of unity: $$\sum_{r=1}^{11} e^{i\frac{2r\pi}{11}} = 0 \quad \Rightarrow \quad \sum_{r=1}^{10} e^{i\frac{2r\pi}{11}} + e^{i2\pi} = 0$$ Since $e^{i2\pi} = 1$, the partial sum simplifies to: $$\sum_{r=1}^{10} e^{i\frac{2r\pi}{11}} = -1 \quad \Rightarrow \quad S = -i(-1) = i$$

Step 2: Set up the outer Arithmetico-Geometric Progression.
Substitute $S = i$ back into the primary expression, letting it be defined as $E$: $$E = \sum_{K=1}^{32} (3K+2)i^K = 5i + 8i^2 + 11i^3 + 14i^4 + \dots + 98i^{32}$$

Step 3: Solve the AGP using the common ratio shift method.
Multiply the entire series expression by the common ratio $i$: $$iE = 5i^2 + 8i^3 + 11i^4 + \dots + 95i^{32} + 98i^{33}$$ Subtracting $iE$ from $E$ shifts the terms to isolate the arithmetic difference $\Delta = 3$: $$E(1-i) = 5i + 3i^2 + 3i^3 + 3i^4 + \dots + 3i^{32} - 98i^{33}$$ $$E(1-i) = 2i + 3\left(i + i^2 + i^3 + \dots + i^{32}\right) - 98i^{33}$$

Step 4: Evaluate the cyclic powers of the imaginary unit $i$.
The terms inside the brackets form a sum of 32 consecutive powers of $i$. Since the sum of any 4 consecutive powers of $i$ vanishes ($i + i^2 + i^3 + i^4 = 0$) and 32 is a perfect multiple of 4, this entire group collapses to 0. Also, evaluate the final term using $i^{32} = 1$: $$98i^{33} = 98(i^{32} \cdot i) = 98i$$ Substitute these simplifications back into the expression: $$E(1-i) = 2i + 3(0) - 98i = -96i \quad \Rightarrow \quad E = \frac{-96i}{1-i}$$

Step 5: Rationalize the complex denominator.
Multiply the numerator and denominator by the complex conjugate $(1+i)$: $$E = \frac{-96i(1+i)}{(1-i)(1+i)} = \frac{-96i - 96i^2}{1 - i^2} = \frac{-96i + 96}{2} = 48(1-i)$$