Rút gọn biểu thức $B = {\sin ^3}\dfrac{a}{3} + 3{\sin ^3}\dfrac{a}{{{3^2}}} + {3^2}{\sin ^3}\dfrac{a}{{{3^3}}} + ... + {3^{n - 1}}{\sin ^3}\dfrac{a}{{{3^n}}}$ bằng:
Trả lời bởi giáo viên
$B = {\sin ^3}\dfrac{a}{3} + 3{\sin ^3}\dfrac{a}{{{3^2}}} + {3^2}{\sin ^3}\dfrac{a}{{{3^3}}} + ... + {3^{n - 1}}{\sin ^3}\dfrac{a}{{{3^n}}}$
$ = \dfrac{{3\sin \dfrac{a}{3} - \sin a}}{4} + 3.\dfrac{{3\sin \dfrac{a}{{{3^2}}} - \sin \dfrac{a}{3}}}{4}$ $ + {3^2}.\dfrac{{3\sin \dfrac{a}{{{3^3}}} - \sin \dfrac{a}{{{3^2}}}}}{4} + .... + {3^{n - 1}}.\dfrac{{3\sin \dfrac{a}{{{3^n}}} - \sin \dfrac{a}{{{3^{n - 1}}}}}}{4}$
$ = \dfrac{1}{4}.\left( { - \sin a + 3\sin \dfrac{a}{3} - 3\sin \dfrac{a}{3}} \right.$ $ + {3^2}\sin \dfrac{a}{{{3^2}}} - {3^2}\sin \dfrac{a}{{{3^2}}} + {3^3}\sin \dfrac{a}{{{3^3}}}$\(\left. { - ... - {3^{n - 1}}\sin \dfrac{a}{{{3^{n - 1}}}} + {3^n}\sin \dfrac{a}{{{3^n}}}} \right)\)
$ = \dfrac{1}{4}\left( {{3^n}\sin \dfrac{a}{{{3^n}}} - \sin a} \right)$ $ = \dfrac{{{3^n}\sin \dfrac{a}{{{3^n}}} - \sin a}}{4}$
Hướng dẫn giải:
Sử dụng công thức ${\sin ^3}\alpha = \dfrac{{3\sin \alpha - \sin 3\alpha }}{4}$.