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Question Number 185423 by alcohol last updated on 21/Jan/23

$$\underset{r=1}{\sum ^n}{3}^{r-1}{sin}^3\left(\frac{\theta }{3^r}\right)=?$$

Answered by witcher3 last updated on 21/Jan/23

$${\sin }^3\left(\mathrm{x}\right)=-\frac{1}{4}\left(\frac{{\mathrm{e}}^{3\mathrm{ix}}-{\mathrm{e}}^{-3\mathrm{ix}}-{3\mathrm{e}}^{\mathrm{ix}}+{3\mathrm{e}}^{-\mathrm{ix}}}{2\mathrm{i}}\right)$$$$=-\frac{1}{4}\sin \left(3\mathrm{x}\right)+\frac{3}{4}\sin \left(\mathrm{x}\right)$$$$\mathrm{S}=\underset{\mathrm{r}=1}{\sum ^{\mathrm{n}}}{3}^{\mathrm{r}-1}{\sin }^3\left(\frac{\theta }{3^{\mathrm{r}}}\right)=-\frac{1}{4}(\underset{\mathrm{r}=1}{\sum ^{\mathrm{n}}}{3}^{\mathrm{r}-1}\sin \left(\frac{\theta }{3^{\mathrm{r}-1}}\right)-\underset{\mathrm{r}=1}{\sum ^{\mathrm{n}}}{3}^{\mathrm{r}}\sin \left(\frac{\theta }{3^{\mathrm{r}}}\right))$$$$=-\frac{1}{4}(\sin \left(\theta \right)-3^{\mathrm{n}}\sin \left(\frac{\theta }{3^{\mathrm{n}}}\right))$$

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