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Question Number 7659 by Tawakalitu. last updated on 07/Sep/16

What is the sum of this sequence sin^2 (x) + sin^2 (2x)+ sin^2 (3x) + ... + sin^2 (nx)

$${What}\:{is}\:{the}\:{sum}\:{of}\:{this}\:{sequence} \\ $$$${sin}^{\mathrm{2}} \left({x}\right)\:+\:{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)+\:{sin}^{\mathrm{2}} \left(\mathrm{3}{x}\right)\:+\:...\:+\:{sin}^{\mathrm{2}} \left({nx}\right) \\ $$

Commented by prakash jain last updated on 07/Sep/16

sin^2 nx=(((e^(inx) −e^(−inx) )/(2i)))^2 =((e^(2inx) −e^(−2inx) −2)/(−2)) =−(e^(2inx) /2)+(e^(−2inx) /2)+1 Given sum=Σ_(j=1) ^n (e^(−2ijx) /2)−Σ_(j=1) ^n (e^(2ijx) /2)+Σ_(j=1) ^n 1 first 2 terms are GP =((e^(−2ix) (e^(−2inx) −1))/(2(e^(−2ix) −1)))−((e^(2ix) (e^(2inx) −1))/(2(e^(2ix) −1)))−n

$$\mathrm{sin}^{\mathrm{2}} {nx}=\left(\frac{{e}^{{inx}} −{e}^{−{inx}} }{\mathrm{2}{i}}\right)^{\mathrm{2}} =\frac{{e}^{\mathrm{2}{inx}} −{e}^{−\mathrm{2}{inx}} −\mathrm{2}}{−\mathrm{2}} \\ $$$$=−\frac{{e}^{\mathrm{2}{inx}} }{\mathrm{2}}+\frac{{e}^{−\mathrm{2}{inx}} }{\mathrm{2}}+\mathrm{1} \\ $$$$\mathrm{Given}\:\mathrm{sum}=\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{e}^{−\mathrm{2}{ijx}} }{\mathrm{2}}−\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{e}^{\mathrm{2}{ijx}} }{\mathrm{2}}+\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{1} \\ $$$${first}\:\mathrm{2}\:{terms}\:{are}\:{GP} \\ $$$$=\frac{{e}^{−\mathrm{2}{ix}} \left({e}^{−\mathrm{2}{inx}} −\mathrm{1}\right)}{\mathrm{2}\left({e}^{−\mathrm{2}{ix}} −\mathrm{1}\right)}−\frac{{e}^{\mathrm{2}{ix}} \left({e}^{\mathrm{2}{inx}} −\mathrm{1}\right)}{\mathrm{2}\left({e}^{\mathrm{2}{ix}} −\mathrm{1}\right)}−{n} \\ $$

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