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

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)

$$\underset{{m}=\mathrm{1}} {\overset{\mathrm{32}} {\sum}}\left(\mathrm{3}{m}+\mathrm{2}\right)\left(\underset{{n}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\left(\mathrm{sin}\left(\frac{\mathrm{2}{n}\pi}{\mathrm{11}}\right)−\mathrm{icos}\left(\frac{\mathrm{2}{n}\pi}{\mathrm{11}}\right)\right)\right)^{{m}} = \\ $$$$\left(\mathrm{A}\right)\:\mathrm{4}\left(\mathrm{1}−{i}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{12}\left(\mathrm{1}+{i}\right) \\ $$$$\left(\mathrm{C}\right)\:\mathrm{12}\left(\mathrm{1}−{i}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{48}\left(\mathrm{1}−{i}\right) \\ $$

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)

$$\mathrm{sin}\:\left(\frac{\mathrm{2}{n}\pi}{\mathrm{11}}\right)−{i}\mathrm{cos}\:\left(\frac{\mathrm{2}{n}\pi}{\mathrm{11}}\right) \\ $$$$=\frac{\mathrm{1}}{{i}}\left\{{i}\mathrm{sin}\:\left(\frac{\mathrm{2}{n}\pi}{\mathrm{11}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{2}{n}\pi}{\mathrm{11}}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{{i}}\centerdot{e}^{{i}\frac{\mathrm{2}{n}\pi}{\mathrm{11}}} =\frac{\mathrm{1}}{{i}}\centerdot{e}^{\frac{\mathrm{2}\pi{i}}{\mathrm{11}}{n}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\left\{\mathrm{sin}\:\left(\frac{\mathrm{2}{n}\pi}{\mathrm{11}}\right)−{i}\mathrm{cos}\:\left(\frac{\mathrm{2}{n}\pi}{\mathrm{11}}\right)\right\}=\frac{\mathrm{1}}{{i}}\underset{{n}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}{e}^{\frac{\mathrm{2}\pi{i}}{\mathrm{11}}{n}} =\frac{\mathrm{1}}{{i}}\centerdot\frac{{e}^{\frac{\mathrm{2}\pi{i}}{\mathrm{11}}} \left(\mathrm{1}−{e}^{\frac{\mathrm{2}\pi{i}}{\mathrm{11}}\centerdot\mathrm{10}} \right)}{\mathrm{1}−{e}^{\frac{\mathrm{2}\pi{i}}{\mathrm{11}}} } \\ $$$$=\frac{\mathrm{1}}{{i}}\centerdot\frac{{e}^{\frac{\mathrm{2}\pi{i}}{\mathrm{11}}} \left({e}^{\frac{\mathrm{2}\pi{i}}{\mathrm{11}}\centerdot\mathrm{11}} −{e}^{\frac{\mathrm{2}\pi{i}}{\mathrm{11}}\centerdot\mathrm{10}} \right)}{\mathrm{1}−{e}^{\frac{\mathrm{2}\pi{i}}{\mathrm{11}}} }=\frac{\mathrm{1}}{{i}}\centerdot\left(−\mathrm{1}\right)\centerdot{e}^{\frac{\mathrm{2}\pi{i}}{\mathrm{11}}+\frac{\mathrm{2}\pi{i}}{\mathrm{11}}\centerdot\mathrm{10}} ={ie}^{\mathrm{2}\pi{i}} ={i} \\ $$$$\underset{{m}=\mathrm{1}} {\overset{\mathrm{32}} {\sum}}\left(\mathrm{3}{m}+\mathrm{2}\right)\left(\underset{{n}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\left(\mathrm{sin}\left(\frac{\mathrm{2}{n}\pi}{\mathrm{11}}\right)−\mathrm{icos}\left(\frac{\mathrm{2}{n}\pi}{\mathrm{11}}\right)\right)\right)^{{m}} \\ $$$$=\underset{{m}=\mathrm{1}} {\overset{\mathrm{32}} {\sum}}\left(\mathrm{3}{m}+\mathrm{2}\right)\centerdot{i}^{{m}} =\mathrm{3}\underset{{m}=\mathrm{1}} {\overset{\mathrm{32}} {\sum}}{m}\centerdot{i}^{{m}} +\mathrm{2}\underset{{m}=\mathrm{1}} {\overset{\mathrm{32}} {\sum}}{i}^{{m}} \\ $$$$=\mathrm{3}\left({i}+\mathrm{2}{i}^{\mathrm{2}} +\mathrm{3}{i}^{\mathrm{3}} +\mathrm{4}{i}^{\mathrm{4}} +\mathrm{5}{i}^{\mathrm{5}} +\mathrm{6}{i}^{\mathrm{6}} +\mathrm{7}{i}^{\mathrm{7}} +...+\mathrm{32}{i}^{\mathrm{32}} \right) \\ $$$$=\mathrm{3}\left\{{i}\left(\mathrm{1}−\mathrm{3}+\mathrm{5}−\mathrm{7}+...+\mathrm{29}−\mathrm{31}\right)−\left(\mathrm{2}−\mathrm{4}+\mathrm{6}−\mathrm{8}+...+\mathrm{30}−\mathrm{32}\right)\right\} \\ $$$$=\mathrm{3}\left\{{i}\left(−\mathrm{2}\right)×\mathrm{8}−\left(−\mathrm{2}\right)×\mathrm{8}\right\} \\ $$$$=−\mathrm{48}\left({i}−\mathrm{1}\right) \\ $$

Commented by EnterUsername last updated on 24/Apr/21

Thank you Sir

$${Thank}\:{you}\:{Sir} \\ $$

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