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Question Number 74819 by Raxreedoroid last updated on 01/Dec/19
Expand Σ ((4n−1)/3)+(2/3)Σ_(k=1) ^(n−1) cos(120k)
$$\mathrm{Expand}\:\Sigma \\ $$$$\frac{\mathrm{4}{n}−\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}}\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}{cos}\left(\mathrm{120}{k}\right) \\ $$
Commented by mathmax by abdo last updated on 01/Dec/19
let S_n =((4n−1)/3) +(2/3)Σ_(k=1) ^(n−1) cos(120k) first let find W_n (α) =Σ_(k=1) ^(n−1) cos(αk) ⇒W_n (α)=Re(Σ_(k=1) ^(n−1) e^(iαk) ) and Σ_(k=1) ^(n−1) e^(iαk) =Σ_(k=0) ^(n−1) (e^(iα) )^k −1 =((1−e^(inα) )/(1−e^(iα) )) −1 =((1−cos(nα)−isin(nα))/(1−cos(α)−isin(α))) −1 =((2sin^2 (((nα)/2))−2isin(((nα)/2))cos(((nα)/2)))/(2sin^2 ((α/2))−2i sin((α/2))cos((α/2))))−1 =((−isin(((nα)/2))e^((inα)/2) )/(−isin((α/2))e^((iα)/2) ))−1 =((sin(((nα)/2)))/(sin((α/2))))e^(i(((n−1)α)/2))) −1 ⇒W_n (α)=((sin(((nα)/2))cos((((n−1)α)/2)))/(sin((α/2))))−1 Σ_(k=1) ^(n−1) cos(120k) =W_n (120) =((sin(60n)cos(60(n−1))/(sin(60)))−1 ⇒ S_n =((4n−1)/3)+(2/3){((sin(60n)cos(60n−60))/(sin(60)))−1}
$${let}\:{S}_{{n}} =\frac{\mathrm{4}{n}−\mathrm{1}}{\mathrm{3}}\:+\frac{\mathrm{2}}{\mathrm{3}}\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:{cos}\left(\mathrm{120}{k}\right)\:\:\:{first}\:{let}\:{find}\: \\ $$$${W}_{{n}} \left(\alpha\right)\:=\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:{cos}\left(\alpha{k}\right)\:\Rightarrow{W}_{{n}} \left(\alpha\right)={Re}\left(\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:{e}^{{i}\alpha{k}} \right) \\ $$$${and}\:\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:{e}^{{i}\alpha{k}} \:=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({e}^{{i}\alpha} \right)^{{k}} −\mathrm{1}\:=\frac{\mathrm{1}−{e}^{{in}\alpha} }{\mathrm{1}−{e}^{{i}\alpha} }\:−\mathrm{1} \\ $$$$=\frac{\mathrm{1}−{cos}\left({n}\alpha\right)−{isin}\left({n}\alpha\right)}{\mathrm{1}−{cos}\left(\alpha\right)−{isin}\left(\alpha\right)}\:−\mathrm{1} \\ $$$$=\frac{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{n}\alpha}{\mathrm{2}}\right)−\mathrm{2}{isin}\left(\frac{{n}\alpha}{\mathrm{2}}\right){cos}\left(\frac{{n}\alpha}{\mathrm{2}}\right)}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}\right)−\mathrm{2}{i}\:{sin}\left(\frac{\alpha}{\mathrm{2}}\right){cos}\left(\frac{\alpha}{\mathrm{2}}\right)}−\mathrm{1}\:=\frac{−{isin}\left(\frac{{n}\alpha}{\mathrm{2}}\right){e}^{\frac{{in}\alpha}{\mathrm{2}}} }{−{isin}\left(\frac{\alpha}{\mathrm{2}}\right){e}^{\frac{{i}\alpha}{\mathrm{2}}} }−\mathrm{1} \\ $$$$=\frac{{sin}\left(\frac{{n}\alpha}{\mathrm{2}}\right)}{{sin}\left(\frac{\alpha}{\mathrm{2}}\right)}{e}^{{i}\left(\frac{\left.{n}−\mathrm{1}\right)\alpha}{\mathrm{2}}\right)} −\mathrm{1}\:\:\Rightarrow{W}_{{n}} \left(\alpha\right)=\frac{{sin}\left(\frac{{n}\alpha}{\mathrm{2}}\right){cos}\left(\frac{\left({n}−\mathrm{1}\right)\alpha}{\mathrm{2}}\right)}{{sin}\left(\frac{\alpha}{\mathrm{2}}\right)}−\mathrm{1} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:{cos}\left(\mathrm{120}{k}\right)\:={W}_{{n}} \left(\mathrm{120}\right)\:=\frac{{sin}\left(\mathrm{60}{n}\right){cos}\left(\mathrm{60}\left({n}−\mathrm{1}\right)\right.}{{sin}\left(\mathrm{60}\right)}−\mathrm{1}\:\Rightarrow \\ $$$${S}_{{n}} =\frac{\mathrm{4}{n}−\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}}\left\{\frac{{sin}\left(\mathrm{60}{n}\right){cos}\left(\mathrm{60}{n}−\mathrm{60}\right)}{{sin}\left(\mathrm{60}\right)}−\mathrm{1}\right\} \\ $$
Commented by mathmax by abdo last updated on 01/Dec/19
i considered that 120 mean 120 radian if the mesure is in degre α =((120π)/(180)) =((2×6π)/(3×6π)) =((2π)/3) in this case we get S_n =((4n−1)/3) +(2/3){ ((sin(((nπ)/3))cos(((n−1)/3)π))/(sin((π/3))))−1}
$${i}\:{considered}\:{that}\:\mathrm{120}\:{mean}\:\mathrm{120}\:{radian}\:\:{if}\:{the}\:{mesure}\:{is}\:{in}\:{degre} \\ $$$$\alpha\:=\frac{\mathrm{120}\pi}{\mathrm{180}}\:=\frac{\mathrm{2}×\mathrm{6}\pi}{\mathrm{3}×\mathrm{6}\pi}\:=\frac{\mathrm{2}\pi}{\mathrm{3}}\:\:{in}\:{this}\:{case}\:{we}\:{get} \\ $$$${S}_{{n}} =\frac{\mathrm{4}{n}−\mathrm{1}}{\mathrm{3}}\:+\frac{\mathrm{2}}{\mathrm{3}}\left\{\:\frac{{sin}\left(\frac{{n}\pi}{\mathrm{3}}\right){cos}\left(\frac{{n}−\mathrm{1}}{\mathrm{3}}\pi\right)}{{sin}\left(\frac{\pi}{\mathrm{3}}\right)}−\mathrm{1}\right\} \\ $$
Answered by mind is power last updated on 01/Dec/19
cos(kx)=Re(e^(ikx) ) Σ_(k=1) ^(n−1) e^(ikx) =e^(ix) .(((1−(e^(ix) )^(n−1) )/(1−e^(ix) )) =(e^(ix(((n+1)/2))) /e^(i(x/2)) ).((sin(((n−1)/2))x)/(sin((x/2)))),Re(Σe^(ikx) )=Σcos(kx)=cos(((nx)/2)).((sin(((n−1)/2)x))/(sin((x/2)))) forx=((2π)/3) S=cos(((nπ)/3)).((sin((n−1)(π/3)))/(sin((π/3)))) =(2/( (√3))).cos(((nπ)/3))sin(((n−1)/3)π) sin(((n−1)/3)π)=sin(((nπ)/3))(1/2)−((√3)/2)cos(n(π/3)) cos(((nπ)/3)),sin(((nπ)/3)) (e^(i(π/3)) )^n =f(n) We worck modd 6 n≡0(6),f(n)=1,n≡1(6),f(n)=(1/2)+((i(√3))/2) n≡2(6),f(n)=−(1/2)+((i(√3))/2) n≡3(6),f(n)=−1 n≡4(6),f(n)=−(1/2)−((i(√3))/2) n≡5(6),f(n)=(1/2)−i((√3)/2) S=((cos(((nπ)/3))sin(((nπ)/3)))/( (√3)))−cos^2 (((nπ)/3)) put different case n modd 6 get close forme add ((4n−1)/3) ge final answer
$$\mathrm{cos}\left(\mathrm{kx}\right)=\mathrm{Re}\left(\mathrm{e}^{\mathrm{ikx}} \right) \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\mathrm{e}^{\mathrm{ikx}} =\mathrm{e}^{\mathrm{ix}} .\frac{\left(\mathrm{1}−\left(\mathrm{e}^{\mathrm{ix}} \right)^{\mathrm{n}−\mathrm{1}} \right.}{\mathrm{1}−\mathrm{e}^{\mathrm{ix}} } \\ $$$$=\frac{\mathrm{e}^{\mathrm{ix}\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}}\right)} }{\mathrm{e}^{\mathrm{i}\frac{\mathrm{x}}{\mathrm{2}}} }.\frac{\mathrm{sin}\left(\frac{\mathrm{n}−\mathrm{1}}{\mathrm{2}}\right)\mathrm{x}}{\mathrm{sin}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)},\mathrm{Re}\left(\Sigma\mathrm{e}^{\mathrm{ikx}} \right)=\Sigma\mathrm{cos}\left(\mathrm{kx}\right)=\mathrm{cos}\left(\frac{\mathrm{nx}}{\mathrm{2}}\right).\frac{\mathrm{sin}\left(\frac{\mathrm{n}−\mathrm{1}}{\mathrm{2}}\mathrm{x}\right)}{\mathrm{sin}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)} \\ $$$$\mathrm{forx}=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\mathrm{S}=\mathrm{cos}\left(\frac{\mathrm{n}\pi}{\mathrm{3}}\right).\frac{\mathrm{sin}\left(\left(\mathrm{n}−\mathrm{1}\right)\frac{\pi}{\mathrm{3}}\right)}{\mathrm{sin}\left(\frac{\pi}{\mathrm{3}}\right)} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}.\mathrm{cos}\left(\frac{\mathrm{n}\pi}{\mathrm{3}}\right)\mathrm{sin}\left(\frac{\mathrm{n}−\mathrm{1}}{\mathrm{3}}\pi\right) \\ $$$$\mathrm{sin}\left(\frac{\mathrm{n}−\mathrm{1}}{\mathrm{3}}\pi\right)=\mathrm{sin}\left(\frac{\mathrm{n}\pi}{\mathrm{3}}\right)\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\left(\mathrm{n}\frac{\pi}{\mathrm{3}}\right) \\ $$$$\mathrm{cos}\left(\frac{\mathrm{n}\pi}{\mathrm{3}}\right),\mathrm{sin}\left(\frac{\mathrm{n}\pi}{\mathrm{3}}\right) \\ $$$$\left(\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{3}}} \right)^{\mathrm{n}} =\mathrm{f}\left(\mathrm{n}\right) \\ $$$$\mathrm{We}\:\mathrm{worck}\:\mathrm{modd}\:\mathrm{6} \\ $$$$\mathrm{n}\equiv\mathrm{0}\left(\mathrm{6}\right),\mathrm{f}\left(\mathrm{n}\right)=\mathrm{1},\mathrm{n}\equiv\mathrm{1}\left(\mathrm{6}\right),\mathrm{f}\left(\mathrm{n}\right)=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{n}\equiv\mathrm{2}\left(\mathrm{6}\right),\mathrm{f}\left(\mathrm{n}\right)=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{n}\equiv\mathrm{3}\left(\mathrm{6}\right),\mathrm{f}\left(\mathrm{n}\right)=−\mathrm{1} \\ $$$$\mathrm{n}\equiv\mathrm{4}\left(\mathrm{6}\right),\mathrm{f}\left(\mathrm{n}\right)=−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{n}\equiv\mathrm{5}\left(\mathrm{6}\right),\mathrm{f}\left(\mathrm{n}\right)=\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{S}=\frac{\mathrm{cos}\left(\frac{\mathrm{n}\pi}{\mathrm{3}}\right)\mathrm{sin}\left(\frac{\mathrm{n}\pi}{\mathrm{3}}\right)}{\:\sqrt{\mathrm{3}}}−\mathrm{cos}^{\mathrm{2}} \left(\frac{\mathrm{n}\pi}{\mathrm{3}}\right) \\ $$$$\mathrm{put}\:\mathrm{different}\:\mathrm{case}\:\mathrm{n}\:\mathrm{modd}\:\mathrm{6}\:\mathrm{get}\:\mathrm{close}\:\mathrm{forme}\:\mathrm{add}\:\frac{\mathrm{4n}−\mathrm{1}}{\mathrm{3}}\:\:\mathrm{ge}\:\mathrm{final}\:\mathrm{answer} \\ $$
Commented by Raxreedoroid last updated on 01/Dec/19
I dont know why but I got a different solution I got ((2(√3)cos(((2π)/3)n−((5π)/6))+12x−6)/9)
$$\mathrm{I}\:\mathrm{dont}\:\mathrm{know}\:\mathrm{why}\:\mathrm{but}\:\mathrm{I}\:\mathrm{got}\:\mathrm{a}\:\mathrm{different}\:\mathrm{solution} \\ $$$$\mathrm{I}\:\mathrm{got} \\ $$$$\frac{\mathrm{2}\sqrt{\mathrm{3}}{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}{n}−\frac{\mathrm{5}\pi}{\mathrm{6}}\right)+\mathrm{12}{x}−\mathrm{6}}{\mathrm{9}} \\ $$

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