r-1-n-3-r-1-sin-3-3-r-

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

\underset{r = 1}{\sum^{n}} 3^{r - 1} {s i n}^{3} (\frac{θ}{3^{r}}) = ?

Answered by witcher3 last updated on 21/Jan/23

\sin^{3} (x) = - \frac{1}{4} (\frac{e^{3 ix} - e^{- 3 ix} - {3 e}^{ix} + {3 e}^{- ix}}{2 i})

= - \frac{1}{4} \sin (3 x) + \frac{3}{4} \sin (x)

S = \underset{r = 1}{\sum^{n}} 3^{r - 1} \sin^{3} (\frac{θ}{3^{r}}) = - \frac{1}{4} (\underset{r = 1}{\sum^{n}} 3^{r - 1} \sin (\frac{θ}{3^{r - 1}}) - \underset{r = 1}{\sum^{n}} 3^{r} \sin (\frac{θ}{3^{r}}))

= - \frac{1}{4} (\sin (θ) - 3^{n} \sin (\frac{θ}{3^{n}}))

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