2013 RI P1 Q8 [Modified]
(i)
Show that ${{\sin }^{4}}A={{\sin }^{2}}A-\frac{1}{4}{{\sin }^{2}}2A$.
It is given that ${{S}_{n}}=\sum\limits_{r=0}^{n}{\frac{1}{{{4}^{r}}}{{\sin }^{4}}\left[ {{2}^{r}}\left( \frac{\pi }{3} \right) \right]}$.
(ii)
By using the result in (i), prove that ${{S}_{n}}=\frac{3}{4}-\frac{1}{{{4}^{n+1}}}{{\sin }^{2}}\left[ {{2}^{n+1}}\left( \frac{\pi }{3} \right) \right]$.
(iii)
Hence give a reason why ${{S}_{n}}$ converges and state the sum to infinity.
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