Σ_(m=1) ^(32) (3m+2)(Σ_(n=1) ^(10) (sin(((2nπ)/(11)))−icos(((2nπ)/(11)))))^m = (A) 4(1−i) (B) 12(1+i) (C) 12(1−i) (D) 48(1−i)


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Question Number 139193 by EnterUsername last updated on 23/Apr/21

$\underset{m = 1}{\sum^{32}} (3 m + 2) {(\underset{n = 1}{\sum^{10}} (\sin (\frac{2 n π}{11}) - icos (\frac{2 n π}{11})))}^{m} =$ $(A) 4 (1 - i) (B) 12 (1 + i)$ $(C) 12 (1 - i) (D) 48 (1 - i)$
	Answered by qaz last updated on 24/Apr/21
	$sin (((2nπ)/(11)))−icos (((2nπ)/(11))) =(1/i){isin (((2nπ)/(11)))+cos (((2nπ)/(11)))} =(1/i)∙e^(i((2nπ)/(11))) =(1/i)∙e^(((2πi)/(11))n) Σ_(n=1) ^(10) {sin (((2nπ)/(11)))−icos (((2nπ)/(11)))}=(1/i)Σ_(n=1) ^(10) e^(((2πi)/(11))n) =(1/i)∙((e^((2πi)/(11)) (1−e^(((2πi)/(11))∙10) ))/(1−e^((2πi)/(11)) )) =(1/i)∙((e^((2πi)/(11)) (e^(((2πi)/(11))∙11) −e^(((2πi)/(11))∙10) ))/(1−e^((2πi)/(11)) ))=(1/i)∙(−1)∙e^(((2πi)/(11))+((2πi)/(11))∙10) =ie^(2πi) =i Σ_(m=1) ^(32) (3m+2)(Σ_(n=1) ^(10) (sin(((2nπ)/(11)))−icos(((2nπ)/(11)))))^m =Σ_(m=1) ^(32) (3m+2)∙i^m =3Σ_(m=1) ^(32) m∙i^m +2Σ_(m=1) ^(32) i^m =3(i+2i^2 +3i^3 +4i^4 +5i^5 +6i^6 +7i^7 +...+32i^(32) ) =3{i(1−3+5−7+...+29−31)−(2−4+6−8+...+30−32)} =3{i(−2)×8−(−2)×8} =−48(i−1)$
	$\sin (\frac{2 n π}{11}) - i \cos (\frac{2 n π}{11})$ $= \frac{1}{i} {i \sin (\frac{2 n π}{11}) + \cos (\frac{2 n π}{11})}$ $= \frac{1}{i} \cdot e^{i \frac{2 n π}{11}} = \frac{1}{i} \cdot e^{\frac{2 π i}{11} n}$ $\underset{n = 1}{\sum^{10}} {\sin (\frac{2 n π}{11}) - i \cos (\frac{2 n π}{11})} = \frac{1}{i} \underset{n = 1}{\sum^{10}} e^{\frac{2 π i}{11} n} = \frac{1}{i} \cdot \frac{e^{\frac{2 π i}{11}} (1 - e^{\frac{2 π i}{11} \cdot 10})}{1 - e^{\frac{2 π i}{11}}}$ $= \frac{1}{i} \cdot \frac{e^{\frac{2 π i}{11}} (e^{\frac{2 π i}{11} \cdot 11} - e^{\frac{2 π i}{11} \cdot 10})}{1 - e^{\frac{2 π i}{11}}} = \frac{1}{i} \cdot (- 1) \cdot e^{\frac{2 π i}{11} + \frac{2 π i}{11} \cdot 10} = {i e}^{2 π i} = i$ $\underset{m = 1}{\sum^{32}} (3 m + 2) {(\underset{n = 1}{\sum^{10}} (\sin (\frac{2 n π}{11}) - icos (\frac{2 n π}{11})))}^{m}$ $= \underset{m = 1}{\sum^{32}} (3 m + 2) \cdot i^{m} = 3 \underset{m = 1}{\sum^{32}} m \cdot i^{m} + 2 \underset{m = 1}{\sum^{32}} i^{m}$ $= 3 (i + 2 i^{2} + 3 i^{3} + 4 i^{4} + 5 i^{5} + 6 i^{6} + 7 i^{7} + . . . + 32 i^{32})$ $= 3 {i (1 - 3 + 5 - 7 + . . . + 29 - 31) - (2 - 4 + 6 - 8 + . . . + 30 - 32)}$ $= 3 {i (- 2) \times 8 - (- 2) \times 8}$ $= - 48 (i - 1)$
	Commented by EnterUsername last updated on 24/Apr/21

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