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Question Number 185423 by alcohol last updated on 21/Jan/23
$$\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{3}^{{r}−\mathrm{1}} {sin}^{\mathrm{3}} \left(\frac{\theta}{\mathrm{3}^{{r}} }\right)\:=\:? \\ $$
Answered by witcher3 last updated on 21/Jan/23
$$\mathrm{sin}^{\mathrm{3}} \left(\mathrm{x}\right)=−\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{e}^{\mathrm{3ix}} −\mathrm{e}^{−\mathrm{3ix}} −\mathrm{3e}^{\mathrm{ix}} +\mathrm{3e}^{−\mathrm{ix}} }{\mathrm{2i}}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\left(\mathrm{3x}\right)+\frac{\mathrm{3}}{\mathrm{4}}\mathrm{sin}\left(\mathrm{x}\right) \\ $$$$\mathrm{S}=\underset{\mathrm{r}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{3}^{\mathrm{r}−\mathrm{1}} \mathrm{sin}^{\mathrm{3}} \left(\frac{\theta}{\mathrm{3}^{\mathrm{r}} }\right)=−\frac{\mathrm{1}}{\mathrm{4}}\left(\underset{\mathrm{r}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{3}^{\mathrm{r}−\mathrm{1}} \mathrm{sin}\left(\frac{\theta}{\mathrm{3}^{\mathrm{r}−\mathrm{1}} }\right)−\underset{\mathrm{r}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{3}^{\mathrm{r}} \mathrm{sin}\left(\frac{\theta}{\mathrm{3}^{\mathrm{r}} }\right)\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{sin}\left(\theta\right)−\mathrm{3}^{\mathrm{n}} \mathrm{sin}\left(\frac{\theta}{\mathrm{3}^{\mathrm{n}} }\right)\right) \\ $$