Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{u_n} = \cos \frac{{\left( {3n + 1} \right)\pi }}{6}\\
{u_{n + 4}} = \cos \frac{{\left[ {3\left( {n + 4} \right) + 1} \right]\pi }}{6} = \cos \frac{{\left( {3n + 1} \right)\pi + 12\pi }}{6} = \cos \left[ {\frac{{\left( {3n + 1} \right)\pi }}{6} + 2\pi } \right]\\
= \cos \frac{{\left( {3n + 1} \right)\pi }}{6} = {u_n}\\
{u_1} = - \frac{1}{2};\,\,\,\,{u_2} = - \frac{{\sqrt 3 }}{2};\,\,\,\,{u_3} = \frac{1}{2};\,\,\,\,{u_4} = \frac{{\sqrt 3 }}{2}\\
\Rightarrow {u_1} + {u_2} + {u_3} + {u_4} = 0\\
{S_{27}} = {u_1} + {u_2} + {u_3} + .... + {u_{26}} + {u_{27}}\\
= 6.\left( {{u_1} + {u_2} + {u_3} + {u_4}} \right) + \left( {{u_{25}} + {u_{26}} + {u_{27}}} \right)\\
= 6.0 + \left( {{u_1} + {u_2} + {u_3}} \right) = - \frac{{\sqrt 3 }}{2}
\end{array}\)
