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Question Number 134251 by Dwaipayan Shikari last updated on 01/Mar/21

Σ_(n=1) ^n nsin(n)

$$\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}{nsin}\left({n}\right) \\ $$

Commented by Dwaipayan Shikari last updated on 01/Mar/21

Σ_(n=1) ^n cos(nθ)=(1/(2sin(θ/2)))(sin((3θ)/2)−sin(θ/2)+...+sin((2n+1)/2)θ−sin((2n−1)/2)θ)  =((cos(((n+1)/2)θ)sin((nθ)/2))/(sin(θ/2)))  −Σ_(n=1) ^n nsin(nθ)=(d/dθ)(((cos(((n+1)/2)θ)sin((nθ)/(2 )))/(sin(θ/2))))  =((sin(θ/2)(−((n+1)/2)sin((nθ)/2)sin(((n+1)/2)θ))−(1/2)cos(θ/2)cos((n+1)/2)θsin((nθ)/2))/(sin^2 (θ/2)))   Σ_(n=1) ^n nsin(nθ)=(((n+1)sin(θ/2)sin((nθ)/2)sin((n+1)/2)θ+cos(θ/2)cos((n+1)/2)θsin((nθ)/2))/(2sin^2 (θ/2)))

$$\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}{cos}\left({n}\theta\right)=\frac{\mathrm{1}}{\mathrm{2}{sin}\frac{\theta}{\mathrm{2}}}\left({sin}\frac{\mathrm{3}\theta}{\mathrm{2}}−{sin}\frac{\theta}{\mathrm{2}}+...+{sin}\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}}\theta−{sin}\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}}\theta\right) \\ $$$$=\frac{{cos}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\theta\right){sin}\frac{{n}\theta}{\mathrm{2}}}{{sin}\frac{\theta}{\mathrm{2}}} \\ $$$$−\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}{nsin}\left({n}\theta\right)=\frac{{d}}{{d}\theta}\left(\frac{{cos}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\theta\right){sin}\frac{{n}\theta}{\mathrm{2}\:}}{{sin}\frac{\theta}{\mathrm{2}}}\right) \\ $$$$=\frac{{sin}\frac{\theta}{\mathrm{2}}\left(−\frac{{n}+\mathrm{1}}{\mathrm{2}}{sin}\frac{{n}\theta}{\mathrm{2}}{sin}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\theta\right)\right)−\frac{\mathrm{1}}{\mathrm{2}}{cos}\frac{\theta}{\mathrm{2}}{cos}\frac{{n}+\mathrm{1}}{\mathrm{2}}\theta{sin}\frac{{n}\theta}{\mathrm{2}}}{{sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}\: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}{nsin}\left({n}\theta\right)=\frac{\left({n}+\mathrm{1}\right){sin}\frac{\theta}{\mathrm{2}}{sin}\frac{{n}\theta}{\mathrm{2}}{sin}\frac{{n}+\mathrm{1}}{\mathrm{2}}\theta+{cos}\frac{\theta}{\mathrm{2}}{cos}\frac{{n}+\mathrm{1}}{\mathrm{2}}\theta{sin}\frac{{n}\theta}{\mathrm{2}}}{\mathrm{2}{sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}\: \\ $$

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